J.   Henry  Senger 


a,:1/' 


WENTWORTH  &  HILL'S 


EXERCISE    MANUALS, 


No.  III. 


GEOMETRY. 


BOSTON: 

GINN,   HEATH,    &   COMPANY. 

1884. 


W4 


Entered,  according  to  Act  of  Congress,  in  the  year  1883, 

BY  GEORGE  A.  WENTWORTH  AND  GEORGE  A.  HILL, 

In  the  Office  of  the  Librarian  of  Congress  at  Washington. 


IN  MEMORIAW 

.r  C 


J.  8.  CUSHING  &  Co.,  PBINTEKS,  BOSTON. 


PREFACE. 


aim  of  wisely-directed  mathematical  teaching  is  to  cultivate 
~L  the  reasoning  faculty,  not  the  memory  ;  and  the  true  test  of 
mathematical  training  is  the  power  which  the  learner  has  acquired 
over  original  problems. 

This  truth  is  very  generally  recognized  in  teaching  Arithmetic 
and  Algebra,  but  very  generally  ignored  in  teaching  Geometry. 
There  are,  however,  many  signs  that  a  change  in  the  method  of 
teaching  Geometry  is  taking  place  in  this  country.  Most  of  the 
recent  text-books  of  Geometry  contain  exercises  designed  to  stimu- 
late original  thought;  most  papers  now  set  in  this  subject  for  admis- 
sion to  college  require  more  or  less  original  work  ;  and  intelligent 
teachers  are  demanding  a  collection  of  suitable  geometrical  exercises. 

The  present  work  has  been  prepared  to  meet  this  demand,  with 
the  hope  that  it  will  promote  a  much-needed  reform.  It  is  the 
first  work  in  the  English  language,  so  far  as  the  authors  know,  in 
which  the  subject  of  geometrical  exercises  is  systematically  treated. 
The  materials  for  the  work  have  been  drawn  chiefly  from  French 
and  German  sources.  The  arrangement  and  mode  of  treatment  are 
such  as  seem  best  adapted  to  meet  the  wants  of  American  schools. 

It  is  not  intended  that  each  student  should  try  to  work  every 
exercise  in  the  book,  but  the  teacher  should  assign  particular  prob- 
lems to  separate  students,  selecting  them  with  special  reference 
to  the  capacity  and  skill  of  each  pupil.  With  care  on  the  part  of 
the  teacher,  pupils  will  gain  the  mastery  over  problems  in  Geometry 
as  readily  as  they  do  over  problems  in  Algebra,  and  precisely  in  the 

SS6505 


IV  PREFACE. 


same  way ;  namely,  by  working  them  out.  It  is  necessary  at  first 
to  give  easy  problems  ;  but  the  doing  of  easy  problems  prepares  the 
way  for  harder  ones  and  still  harder. 

The  exercises  here  given  consist  of  a  great  number  of  easy  prob- 
lems for  beginners,  and  enough  harder  ones  for  more  advanced 
scholars.  The  exercises  in  each  section  are  carefully  graded,  and 
some  of  the  more  difficult  sections  can  be  omitted  without  destroying 
the  unity  of  the  work.  The  book  can  be  used  in  connection  with 
any  text-book  on  Geometry,  as  soon  as  the  geometrical  processes  of 
reasoning  are  well  understood. 

A  Syllabus  of  Geometry  is  given,  not  only  for  reference,  but  with 
a  view  of  making  the  book  by  itself  convenient  for  reviewing  the 
study  of  Geometry.  Lessons  can  be  assigned  consisting  partly  of 
book-work  taken  from  the  Syllabus,  and  partly  of  original  work, 
and  the  two  parts  can  be  so  fitted  to  each  other  that  a  thorough 
knowledge  of  the  book-work  will  be  necessary  in  order  to  do  the 
original  work,  and  the  doing  of  the  original  work  will  firmly  fix 
in  the  mind  the  principles  involved  in  the  book-work. 

Any  corrections  or  any  suggestions  relating  to  the  work  will  be 

thankfully  received. 

G.  A.  WENTWORTH. 

G.  A.  HILL. 
PHILLIPS  EXETER  ACADEMY, 
September,  1884. 


CONTENTS. 


PLANE    GEOMETRY 

CHAPTER  I.     THE  STRAIGHT  LINE. 

PAOB 

Questions  and  Numerical  Exercises          ....  1 

Theorems 5 

CHAPTER  II.     THE  CIRCLE. 

Questions  and  Numerical  Exercises          ....  10 

Theorems         .........  14 

Loci 21 

Problems;  General  Remarks 26 

Construction  of  Points 28 

Construction  of  Circles     .......  31 

Construction  of  Straight  Lines 35 

Construction  of  Triangles         ......  44 

Construction  of  Quadrilaterals 53 

Miscellaneous  Exercises  .......  57 

CHAPTER  III.     SIMILAR  FIGURES. 

Theorems 60 

Numerical  Exercises         .......  70 

Loci 79 

Problems 86 

The  Method  of  Similar  Figures 92 

The  Problem  of  Apollonius .97 

CHAPTER   IV.     EQUIVALENT  FIGURES. 

Theorems 102 

Numerical  Exercises         .......  105 

Problems  119 


VI  CONTENTS. 


CHAPTER  V.     REGULAR  FIGURES. 

PAGE 

Theorems 130 

Numerical  Exercises 132 

Problems          .        . 146 

CHAPTER  VI.    THE  ALGEBRAIC  METHOD. 

Construction  of  Algebraic  Expressions    .        .        .        .151 

Homogeneity  of  Algebraic  Expressions  ....  156 

Examples  of  the  Algebraic  Method          ....  158 

Classified  Equations  of  the  First  Degree  ....  164 

Unclassified  Equations  of  the  First  Degree      .        .         .  166 

Pure  Quadratic  Equations       .         .        .        .        .         .  169 

Complete  Quadratic  Equations 170 


SOLID    GEOMETRY. 

Planes 1 

The  Prism 4 

The  Cylinder 7 

The  Pyramid 10 

The  Cone 13 

Frustums  of  Pyramids  and  Cones 16 

The  Sphere 19 

Equivalent  Solids 27 

Similar  Solids 29 

Solids  of  Revolution  .         .         .         .         .  .         .         .32 

Inscribed  and  Circumscribed  Solids 36 

Useful  Formulae .                                                                             .  38 


SYLLABUS 


OF 


PLANE   GEOMETRY. 


AXIOMS. 

1.  Magnitudes  which  can  be  made  to  coincide  are  equal. 

2.  Two  magnitudes,  each  equal  to  a  third,  are  equal  to 
each  other. 

3.  If  equals  are  added  to  equals,  the  sums  are  equal. 

4.  If  equals  are  taken  from  equals,  the  remainders  are 
equal. 

5.  If  equals  are  added  to  unequals,  the  sums  are  unequal 
in  the  same  sense. 

6.  If  equals  are  taken  from  unequals,  the  remainders  are 
unequal  in  the  same  sense. 

7.  If  unequals  are  taken  from  equals,  the  remainders  are 
unequal  in  the  opposite  sense. 

8.  The  whole  is  equal  to  the  sum  of  its  parts. 

9.  Through  two  points  only  one  straight  line  can   be 
drawn. 

10.  A  straight   line  is  the  shortest  line  between  two 
points. 

11.  Through  a  point  not  in  a  straight  line  only  one  par- 
allel to  the  line  can  be  drawn. 


Vlll  GEOMETRY. 


;    '         BOOK  I. 


THE  STRAIGHT  LINE. 


DEFINITIONS. 

12.  Body,  surface,  line,  point,  straight  line,  curved  line, 
broken  line,  plane  surface  or  plane,  curved  surface,  figure, 
plane  figure,  similar  figures,  equivalent  figures,  equal  fig- 
ures, test  of  the  equality  of  geometrical  figures,  method  of 
superposition. 

Object  of  Geometry,  Plane  Geometry,  Solid  Geometry, 
axiom,  theorem,  corollary,  scholium,  problem,  postulate, 
proposition,  hypothesis,  conclusion,  proof,  converse  theo- 
rem, contrary  theorem. 

Comparison  of  lines  as  regards  magnitude,  linear  units, 
length  of  a  line,  addition,  subtraction,  multiplication,  and 
division  of  lines. 

Angle,  its  sides,  its  vertex,  naming  of  an  angle,  straight 
angle,  right  angle,  acute  angle,  obtuse  angle,  generation  of 
an  angle,  comparison  of  angular  magnitudes,  division  of  the 
right  angle  into  degrees,  minutes,  and  seconds,  adjacent 
angles,  vertical  angles,  supplementary  angles,  complemen- 
tary angles,  bisector  of  an  angle,  perpendicular  lines, 
oblique  lines. 

Parallel  lines,  secant,  alternate-interior  angles,  alternate- 
exterior  angles,  exterior-interior  angles,  interior  and  exte- 
rior angles  on  the  same  side  of  the  secant. 

Polygon,  its  sides,  its  angles,  its  vertices,  its  parts,  tri- 
angle, quadrilateral,  pentagon,  hexagon,  octagon,  decagon, 
dodecagon,  perimeter,  diagonal,  exterior  angles,  convex 


SYLLABUS.  IX 


polygon,  reentrant  angles,  equilateral  polygon,  equiangular 
polygon,  mutually  equiangular  polygons,  mutually  equi- 
lateral polygons,  homologous  vertices,  angles,  and  sides, 
equal  polygons. 

Equilateral  triangle,  isosceles  triangle,  scalene  triangle, 
right  triangle,  acute  triangle,  obtuse  triangle,  hypotenuse, 
legs,  base,  vertex,  altitude. 

A  triangle  has  three  altitudes  because  each  side  may 
be  taken  as  base.  The  three  lines  drawn  from  the  vertices 
to  the  opposite  sides  and  bisecting  the  angles  are  called  the 
bisectors.  The  three  lines  drawn  from  the  vertices  to  the 
middle  points  of  the  opposite  sides  are  called  the  medi- 
ans. In  the  isosceles  triangle  the  two  equal  sides  are 
called  the  legs  ;  the  other  side  is  called  the  base ;  and  the 
vertex  of  the  angle  opposite  to  the  base  is  called  the  vertex 
of  the  triangle. 

Parallelogram,  square,  rectangle,  rhombus,  rhomboid, 
trapezoid. 

The  two  parallel  sides  of  a  trapezoid  are  called  the  bases ; 
the  other  sides,  the  legs;  the  line  joining  the  middle  points 
of  the  legs,  the  median.  An  isosceles  trapezoid  is  a  trape- 
zoid having  equal  legs. 

A  deltoid  is  a  quadrilateral  having  two  adjacent  sides 
equal,  and  the  other  two  sides  also  equal. 

The  centre  of  a  parallelogram  is  the  intersection  of  its 
diagonals. 

ANGLES,  PERPENDICULARS,  AND  PARALLELS. 

13.  Theorem.   All  straight  angles  are  equal. 

14.  Corollary  I.    All  right  angles  are  equal. 

15.  Corollary  II.    The  sum  of  the  adjacent  angles  formed 
by  producing  one  side  of  an  angle  from  the  vertex  is  equal 
to  180°. 


GEOMETRY. 


16.  Corollary  III.    The  sum  of  all  the  consecutive  angles 
which  can  be  formed  on  one  side  of  a  straight  line  at  the 
same  point  in  the  line  is  equal  to  180°. 

17.  Corollary  IV.    The  sum  of  all  the  consecutive  angles 
which  can  be  formed  about  a  point  is  equal  to  360°. 

18.  Theorem,   If  the   sum   of  two   adjacent   angles   is 
equal  to  180°,  their  exterior  sides  form  a  straight  line. 

19.  Theorem.    Vertical  angles  are  equal. 

20.  Theorem.   Only  one  perpendicular  can  be  erected  at 
a  given  point  in  a  straight  line. 

21.  Theorem.   Only  one  perpendicular   can   be   drawn 
from  a  given  point  to  a  straight  line. 

22.  Theorem,.    Two  straight  lines,  each  perpendicular  to 
the  same  straight  line,  are  parallel. 

23.  Theorem.    If  a  straight  line  is  perpendicular  to  one 
of  two  parallels,  it  is  also  perpendicular  to  the  other. 

24.  Theorem..    If  two  straight  lines  are  each  parallel  to 
a  third  line,  they  are  parallel  to  each  other. 

25.  Theorem.    If  two  parallels  are  cut  by  a  secant : 

(i.)    the  alternate-interior  angles  are  equal ; 
(ii.)    the  alternate-exterior  angles  are  equal ; 
(iii.)   the  exterior-interior  angles  are  equal ; 
(iv.)   the  interior  or  the  exterior  angles  on  the  same 
side  of  the  secant  are  supplementary. 

26.  Theorem.  If  any  one  of  the  four  conditions  in  No.  25 
is  satisfied,  the  two  lines  cut  by  the  secant  are  parallel. 

27.  Theorem.    Two  angles  whose  sides  are  respectively 
parallel  are  equal   or  supplementary :    equal,  if  both  are 
acute  or  both  obtuse;  supplementary,  if  one  is  acute  and 
the  other  obtuse. 

28.  Theorem.    Two  angles  whose  sides  are  respectively 
perpendicular  are  equal  or  supplementary :  equal,  if  both 
are  acute  or  both  obtuse ;  supplementary,  if  one  is  acute 
and  the  other  obtuse. 


SYLLABUS. 


29.  Theorem.    The  sum  of  the  three  angles  of  a  triangle 
is  equal  to  180°. 

30.  Corollary  I.    If  two  triangles  have  two  angles  equal, 
each  to  each,  the  third  angles  are  also  equal. 

31.  Corollary  II.    The  sum  of  the  acute  angles  of  a  right 
triangle  is  equal  to  90°. 

32.  Theorem.    An  exterior  angle  of  a  triangle  is  equal 
to  the  sum  of  the  two  opposite  interior  angles. 

33.  Theorem.    The  sum  of  the  angles  of  a  quadrilateral 
is  equal  to  360°. 

34.  Theorem.    The  sum  of  the  angles  of  a  polygon  of  n 
sides  is  equal  to  (n  —  2)  X  180°. 

35.  TJieorem.    The  sum  of  the  exterior  angles  of  a  con- 
vex polygon  of  n  sides,  formed  by  producing  each  side  in 
succession,  is  equal  to  360°. 

TRIANGLES. 

36.  Theorem.    Two  triangles  are  equal  if  they  have  a 
side  and  the  two  adjacent  angles  equal,  each  to  each. 

37.  Corollary.    Two  right  triangles  are    equal   if  they 
have  a  side  and  an  acute  angle  of  the  one  equal  to  a  corre- 
sponding side  and  acute  angle  of  the  other. 

38.  Theorem.    Two  triangles  are  equal  if  they  have  two 
sides  and  the  included  angle  equal,  each  to  each. 

39.  Corollary.    Two  right  triangles  are   equal   if  their 
legs  are  equal,  each  to  each. 

40.  Theorem.    In  an  isosceles  triangle  the  angles  oppo- 
site the  equal  sides  are  equal. 

41.  Corollary.    An  equilateral  triangle  is  equiangular. 

42.  Theorem.    If  in  a  triangle  two  angles  are  equal,  the 
opposite  sides  are  equal,  and  the  triangle  is  isosceles. 

43.  Corollary.    An    equiangular   triangle   is   also    equi- 
lateral. 


Xll  GEOMETRY. 


44.  Theorem.    In  an  isosceles  triangle  the  median  drawn 
from  the  vertex  to  the  base  is  perpendicular  to  the  base 
and  bisects  the  angle  at  the  vertex. 

45.  Theorem.     Two  triangles  are  equal  if  their  three 
sides  are  equal,  each  to  each. 

46.  Theorem.      Two    right  triangles  are   equal  if  they 
have  a  leg  and  the  hypotenuse  equal,  each  to  each. 

47.  Theorem.    Of  two  angles  of  a  triangle  that  is  the 
greater  which  is  opposite  the  greater  side. 

48.  Theorem.    Of  two  sides  of   a  triangle  that  is  the 
greater  which  is  opposite  the  greater  angle. 

49.  Theorem.   If  two  triangles  have  two  sides  equal,  each 
to  each,  but  the  included  angles  unequal,  the  third  side  is 
greater  in  that  triangle  which  has  the  greater  angle. 

50.  Theorem.   If  two  triangles  have  two  sides  equal,  each 
to  each,  but  the  third  sides  unequal,  the  angle  included  by 
the  equal  sides  is  greater  in  that  triangle  which  has  the 
greater  side. 

PARALLELOGRAMS. 

51.  Theorem.    The  diagonal  of  a  parallelogram  divides 
it  into  two  equal  triangles. 

52.  Corollary  I.    The  opposite  sides   and   the   opposite 
angles  of  a  parallelogram  are  equal. 

53.  Corollary  II.    Parallels  between  parallels  are  equal. 

54.  Corollary   III.    Parallels    are    everywhere    equally 
distant. 

55.  Theorem.   The  diagonals  of  a  parallelogram  mutu- 
ally bisect  each  other. 

56.  Theorem.    A.  quadrilateral  is  a  parallelogram  if : 

(i.)  the  opposite  sides  are  equal ; 

(ii.)  the  opposite  angles  are  equal; 

(iii.)  two  opposite  sides  are  equal  and  parallel ; 

(iv.)  the  diagonals  mutually  bisect  each  other. 


SYLLABUS.  xiii 


57.  Theorem.    The  diagonals  of  a  rectangle  are  equal. 

58.  Theorem.    The  diagonals  of  a  rhombus  are  perpen- 
dicular to  each  other. 

59.  Corollary.    The  diagonals  of  a  square  are  equal  and 
perpendicular  to  each  other. 

60.  Theorem.    Two    parallelograms   are    equal   if  they 
have  two  adjacent  sides  and  the  included  angle  equal,  each 
io  each. 

THEOREMS  RELATING  TO  LENGTHS  AND  DISTANCES. 

61.  Theorem.    Each  side  of  a  triangle  is  less  than  the 
sum  of  the  other  two  sides,  and  greater  than  their  difference. 

62.  Theorem.    The  sum  of  two  straight  lines,  drawn  from 
a  point  to  the  ends  of  a  straight  line,  is  greater  than 'the 
sum  of  any  other  two  straight  lines  similarly  drawn  but 
included  by  them. 

63.  Theorem,.    A  perpendicular  is  the  shortest  distance 
from  a  point  to  a  straight  line. 

64.  Theorem.    Two  oblique  lines,  drawn  from  a  point  in 
a  perpendicular  so  as  to  cut  off  equal  distances  from  the 
foot  of  the  perpendicular,  are  equal;  and  of  two  oblique 
lines  which  cut  off  unequal  distances,  the  more  remote  is 
the  greater. 

65.  Corollary.    Only  two   equal   straight   lines   can   be 
drawn  from  a  point  to  a  straight  line. 

66.  Theorem.    Two  equal  oblique  lines  cut  off  equal  dis- 
tances from  the  foot  of  the  perpendicular ;  and  of  two  un- 
equal oblique  lines,  the  greater  cuts  off  the  greater  distance 
from  the  foot  of  the  perpendicular. 

67.  Theorem.    Every  point  in  the  perpendicular,  erected 
at  the  middle  point  of  a  straight  line,  is  equidistant  from 
the  ends  of  the  line,  and  every  point  not  in  the  perpendicu- 
lar is  unequally  distant  from  the  ends  of  the  line. 


XIV  GEOMETRY. 


68.  Corollary.    Two  points  equidistant  from  the  ends  of 
a  straight  line  determine  the  perpendicular  through  the 
middle  point  of  the  line. 

69.  Theorem.    Every  point  in  the  bisector  of  an  angle  is 
equidistant  from  the  sides  of  the  angle,  and  every  point 
not  in  the  bisector  is  unequally  distant  from  the  sides. 

70.  Theorem.    The  line  joining  the  middle  points  of  two 
sides  of  a  triangle  is  parallel  to  the  third  side  and  equal  to 
half  of  the  third  side. 

71.  Theorem.   A  straight  line  drawn  parallel  to  the  base 
of  a  triangle,  bisecting  one  of  the  sides,  bisects  the  other 
also ;  and  the  part  intercepted  between  the  two  sides  is 
equal  to  half  the  base. 

72.  Theorem.    The  median  of  a  trapezoid  is  parallel  to 
the  bases  and  equal  to  half  their  sum. 

73.  Theorem.    Equidistant  parallels  divide  every  secant 
into  equal  parts. 


BOOK   II. 

THE  CIRCLE. 


THE  CIRCLE  AND  STRAIGHT  LINES. 

74.  Definitions.  Circumference,  circle,  radius,  diameter, 
arc,  chord,  semi-circumference,  segment,  sector,  quadrant, 
semicircle,  secant,  tangent,  point  of  contact,  chord  of  con- 
tact, inscribed  polygon,  circumscribed  circle,  circumscribed 
polygon,  inscribed  circle. 

A  chord  quadrilateral  is  a  quadrilateral  about  which  a 
circle  can  be  circumscribed. 

A  tangent  quadrilateral  is  a  quadrilateral  in  which  a 
circle  can  be  inscribed. 


SYLLABUS.  XV 


75.  Theorem.    A  straight  line  cannot  meet  a  circumfer- 
ence in  more  than  two  points. 

76.  Theorem.   Every  diameter  divides  a  circle  and  its 
circumference  into  two  equal  parts. 

77.  Theorem.    Every  chord  is  less  than  a  diameter. 

78.  Theorem.    Circles  having  equal  radii  or  equal  diam- 
eters are  equal ;  and  conversely. 

79.  Theorem.  In  the  same  circle,  or  in  equal  circles,  equal 
arcs  are  subtended  by  equal  chords ;  and  conversely. 

80.  Theorem.    The  radius  perpendicular  to  a  chord  bi- 
sects the  chord  and  the  arc  subtended  by  the  chord. 

81.  Theorem.    The  perpendicular  erected  at  the  middle 
point  of  a  chord  passes  through  the  centre  and  bisects  the 
arc  subtended  by  the  chord. 

82.  Scholium.  The  line  mentioned  in  the  last  two  propo- 
sitions satisfies  four  conditions :  (i.)  it  passes  through  the 
centre ;  (ii.)  it  is  perpendicular  to  the  chord ;  (iii.)  it  bi- 
sects the  chord ; .  (iv.)  it  bisects  the  arc  subtended  by  the 
chord.     Any  two  of  these  conditions  determine  this  line. 
Since  two  things  can  be  selected  from  four  things  in  six 
different  ways,  there  are  altogether  six  different  theorems 
respecting  this  line. 

83.  Theorem.    Through  three  points  not  in  a  straight 
line  one  circumference  can  be  drawn,  and  only  one. 

84.  Corollary.    If  two  circumferences  have  three  points 
common,  they  coincide  throughout. 

85.  Theorem.   In   the  same   circle,  or  in  equal  circles, 
equal  chords  are  equally  distant  from  the  centre ;  and  con- 
versely. 

86.  Theorem.    In  the  same  circle,  or  in  equal  circles,  of 
two  unequal  chords  the  greater  is  nearer  the  centre ;  and 
conversely. 

87.  Theorem.    A  tangent  to  a  circle  is  perpendicular  to 
the  radius  drawn  to  the  point  of  contact. 


XVI  GEOMETRY. 


88.  Theorem.    A  perpendicular  to  a  tangent  erected  at 
the  point  of  contact  passes  through  the  centre. 

89.  Theorem.    The  tangents  to  a  circle  drawn  through 
an  exterior  point  are  equal,  and  make  equal  angles  with  the 
line  joining  the  point  to  the  centre. 

VARIABLES  AND  THEIR  LIMITS. 

90.  Definitions.    Measure  .of  a  quantity,  common  meas- 
ure of  two  quantities,  ratio  of  two  quantities,  antecedent, 
consequent,    commensurable    and   incommensurable  ratios, 
approximate  value,  constant,  variable,  limit  of  a  variable. 

91.  Proposition  I.    If  two  variables  are  constantly  equal, 
and  tend  each  towards  a  limit,  these  limits  are  also  equal. 

92.  Proposition  II.    The  limit  of  the  algebraic  sum  of 
two  or  more  variables  is  equal  to  the  algebraic  sum  of  their 
limits. 

93.  Proposition  III.    The  limit  of  the  product  of  two  or 
more  variables  is  equal  to  the  product  of  their  limits. 

94.  Proposition  IV.    The  limit  of  the  ratio  of  two  varia- 
bles is  equal  to  the  ratio  of  their  limits. 

i 
THE  CIRCLE  AND  ANGLES. 

95.  Definitions.    Angle   at  the  centre,  inscribed  angle, 
angle  standing  on  a  chord,  angle  inscribed  in  a  segment. 

96.  Theorem.    In  the  same   circle,  or  in   equal   circles, 
equal  angles  at  the  centre  intercept  equal  arcs ;  and  con- 
versely. 

97.  Theorem.    In  the  same  circle,  or  in  equal  circles, 
the  ratio  of  two  angles  at  the  centre  is  equal  to  the  ratio  of 
the  intercepted  arcs. 

98.  Corollary  I.    An  angle  has  the  same  numerical  meas- 
ure as  an  arc  described  from  its  vertex  as  centre  and  inter- 
cepted by  its  sides,  if  the  unit  of  arc  is  defined  to  be  the 
arc  intercepted  by  the  sides  of  unit  angle. 


SYLLABUS.  xvii 


99.  Theorem.    An  inscribed  angle  is  measured  by  half 
the  intercepted  arc. 

100.  Corollary  I.    All  angles  inscribed  in  the  same  seg- 
ment are  equal. 

101.  Corollary  II.    Every  angle  inscribed  in  a  semicircle 
is  a  right  angle. 

102.  Corollary  III.    Every  angle  inscribed  in  a  segment 
greater  than  a  semicircle  is  acute,  and  every  angle  inscribed 
in  a  segment  less  than  a  semicircle  is  obtuse. 

103.  Theorem.   The  angle  formed  by  a  tangent  and  a 
chord  is  measured  by  half  the  intercepted  arc. 

104.  Theorem.    An  angle  which   has  its  vertex  within 
the  circle  is  measured  by  half  the  sum  of  the  intercepted 
arcs. 

105.  Theorem.    An  angle  which  has  its  vertex  without 
the  circle  is  measured  by  half  the  difference  of  the  inter- 
cepted arcs. 

106.  TJieorem.   The  opposite  angles  of  an  inscribed  quad- 
rilateral are  supplementary. 

107.  Theorem.    If  two  opposite  angles  of  a  quadrilateral 
are  supplementary,  the  quadrilateral  is  a  chord  quadri- 
lateral. 

Two  CIRCLES. 

108.  Definitions.    Concentric  circles,  line  of  centres,  ex- 
ternal contact,  internal  contact,  exterior  tangents,  interior 
tangents. 

109.  Theorem.   If  two  circles  cut  each  other,  the  line  of 
centres  is  perpendicular  to  the  common  chord  and  bisects 
it. 

110.  Theorem.    If  two  circles  touch  each  other,  the  line 
of  centres  passes  through  the  point  of  contact. 


XV111  GEOMETRY. 


PROBLEMS. 

111.  Instruments.    Ruler,    compasses,    set-square,    pro- 
tractor. 

A  problem  does  not  belong  to   Elementary   Geometry 
unless  it  can  be  solved  with  ruler  and  compasses  only. 

112.  To  erect  a  perpendicular  at  a  given  point  in  a  given 
line. 

113.  To  drop  a  perpendicular  from  a  given  point  to  a 
given  line. 

114.  To  bisect  a  given  line. 
11.5.    To  bisect  a  given  arc. 

116.  To  bisect  a  given  angle. 

117.  At  a  given  point  in  a  given  line  to  construct  an 
angle  equal  to  a  given  angle. 

118.  Through  a  given  point  to  draw  a  parallel  to  a  given 
line. 

119.  From  a  point  without  a  given  line  to  draw  a  line 
making  a  given  angle  with  this  line. 

120.  Two  angles  of  a  triangle  being  given,  to  find  the 
third  angle. 

121.  To  construct  a  triangle,  given  two  sides  and  the 
included  angle. 

122.  To  construct  a  triangle,  given  one  side  and  the  two 
adjacent  angles.     When  is  the  problem  impossible? 

123.  To  construct  a  triangle,  given  one  side,  one  adjacent 
angle,  and  the  opposite  angle. 

124.  To    construct   a   triangle,    given    the    three   sides. 
When  is  the  problem  impossible  ? 

125.  To  construct  a  triangle,  given  two  sides  and  the 
angle  opposite  one  of  these  sides.     When  will  the  problem 
have  two  solutions  ?  one  solution  ?  no  solution  ? 

126.  To  divide  a  given  line  into  any  number  of  equal 
parts. 


SYLLABUS.  XIX 


127.  To    find    the    ratio    of  two    given    commensurable 
straight  lines. 

128.  To  find  the  centre  of  a  given  circumference.. 

129.  To  describe  a  circumference  through  three  given 
points;  or,  to  circumscribe  a  circle  about  a  given  triangle. 

130.  To  inscribe  a  circle  in  a  given  triangle. 

131.  Through  a  given  point  to  draw  a  tangent  to  a  given 
circle.   When  will  the  problem  have  one  solution?  two  solu- 
tions ?  no  solution  ? 

132.  To  draw  a  common  tangent  to  two  given  circles. 
When  will  the  problem  have  four  solutions  ?  three  solutions  ? 
two  solutions  ?  one  solution  ?  no  solution  ?. 

133.  To  construct  upon  a  given  line  a  segment  capable 
of  containing  a  given  angle. 


BOOK  III. 

SIMILAR  FIGURES. 


THEORY  OF  PROPORTIONS. 

134.    Definitions;  Proportion,  terms  6f  a  proportion,  ex- 
tremes and  means,. inverse  or  reciprocal  proportion.  , 
.  135.    Theorem.    If  a  :  b  =  c  :  d,  then  ad  =  be. 

136.  Theorem.   If  ad  =  be,  then  a  :  b  ===  c  :  d. 

137.  Theorems.   Ifa:b  =  c:d,  then : 
(i.)  a  :  c  =  b  :  d; 

(ii.)  a±c  :b±d  =  a:b  =  c:d. 
(iii.)  a±b  :  c  ±  d  =  a:  c  —  b  :  d] 
(iv.)  a-\-c:a  —  c  =  b  +  d:b  —  d\ 
(v.)  a  +  b  :a  —  b  =  c-\-d:c  —  d; 
(vi.)  .ma  :  mb  =  nc  :  nd ; 
(vii.)  an  :  bn  =  cn  :  dn. 


XX  GEOMETRY. 


PROPORTIONAL  LINES, 

138.  Definitions,    Proportional  lines,  third  proportional, 
mean  proportional,  extreme  and  mean  ratio,  segments  of  a 
line. 

139.  Theorem.    A  line  parallel  to  one  side  of  a  triangle 
divides  the  other  two   sides  into  proportional  parts  ;  and 
these  two  sides  have  the  same  ratio  as  two  corresponding 
segments. 

140.  Theorem.    A  line  which  divides  two  sides  of  a  tri- 
angle into  proportional  parts  is  parallel  to  the  third  side. 

141.  Theorem.  Parallels  cut  proportional  parts  from  any 
two  straight  lines. 

142.  Theorem.    Lines  meeting  in  a  common  point  divide 
parallels  into  proportional  parts. 

143.  Theorem.   The  bisector  of  an  angle  of  a  triangle 
divides  the  opposite  side  into  segments  proportional  to  the 
adjacent  sides. 

144.  Theorem.   The  bisector  of  an  exterior  angle  of  a 
triangle  divides  the  opposite  side  produced  into  segments 
proportional  to  the  other  two  sides. 

SIMILAR.  TRIANGLES, 

145.  Definitions.   Similar  triangles,  homologous   angles, 
vertices,  sides,  bases,  and  altitudes. 

146.  Theorem.  Two  triangles  are  similar  if  they  are  equi- 
angular with  respect  to  each  other. 

147.  Corollary  I.    Two  triangles  are  similar  if  they  have 
two  angles  equal,  each  to  each. 

148.  Corollary  II.    Two  right  triangles  are  similar  if  an 
acute  angle  of  one  is  equal  to  an  acute  angle  of  the  other. 

149.  Theorem.   Two  triangles  are  similar  if  they  have 
an  angle  of  one  equal  to  an  angle  of  the  other  and  the  in- 
cluding sides  proportional. 


SYLLABUS. 


150.  Corollary.   Two  right  triangles  are  similar  if  their 
legs  are  proportional. 

151.  Theorem.   Two  triangles  are  similar  if  their  sides, 
taken  in  order,  are  proportional. 

152.  Theorem.    The  homologous  altitudes  of  two  similar 
triangles  are  proportional  to  any  two  homologous  sides. 

SIMILAR  POLYGONS. 

153.  Definitions.   Similar  polygons,  homologous  angles, 
vertices,  sides,  and  diagonals. 

154.  Theorem.    Two  similar  polygons  are  divisible  into 
the  same  number  of  triangles,  similar  each  to  each,  and 
similarly  placed. 

155.  Theorem.    Two   polygons   are   similar  if  they  are 
divisible  into  the  same  number  of  triangles,  similar  each  to 
each,  and  similarly  placed. 

156.  Theorem.   The  perimeters  of  two  similar  polygons 
are  proportional  to  any  two  homologous  sides. 

NUMERICAL  PROPERTIES  OF  FIGURES. 

157.  Definition.    The  projection  of  a  line. 

158.  Theorem.    In  a  right  triangle  : 

(i.)  the  altitude  upon  the  hypotenuse  divides  the  trian- 
gle into  two  right  triangles,  similar  to  the  given  triangle 
and  to  each  other ; 

(ii.)  the  altitude  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse ; 

(iii.)  each  leg  is  a  mean  proportional  between  the  hypot- 
enuse and  the  adjacent  segment; 

(iv.)  the  squares  of  the  legs  are  proportional  to  the  adja- 
cent segments ; 

(v.)  the  product  of  the  two  legs  is  equal  to  the  product 
of  the  hypotenuse  and  the  altitude. 


XX11  GEOMETRY. 


159.  Theorem.    A  perpendicular  dropped  from  a  point 
in  the  circumference  of  a  circle  to  a  .diameter  is  a  mean 
proportional  between  the  segments  of  the  diameter.     . 

160.  Theorem.   A  chord  drawn  from  the  end  of  a  diam- 
eter is  a  mean  proportional  between  the  diameter  and  the 
projection  of  the  chord  upon  the  diameter. 

161 .  Theorem.  In  a  right  triangle  the  square  of  the  hypot- 
enuse is  equal  to  the  sum  of  the  squares  of  the  two  legs. 

1^2.  Theorem  In  a  triangle  the  square  of  a  side  opposite 
an  acute  angle  is  equal  to  the  sum  of  'the  squares  of  the 
other  two  -sides,  diminished  by  twice  the  product  of  one  of 
these  sides  and  the  projection  of  the  other  upon  it. 

163.  Theorem.    In  an  obtuse  triangle  the  square  of  the 
side  opposite  the  obtuse  angle  is  equal-, to  the  sum  of  the 
squares  of  the  other  two  sides,  increased  by  twice  the  product 
of  one  of  these  sides  and  the  projection  of  the  other  upon  it. 

164.  Theorem.    If  through  a  fixed  point  within  a  circle 
a  chord  is  drawn,  the  product  of  the  two  segments  of  the 
chord  is  constant  in  whatever  direction  the  chord  is  drawn. 

165.  Theorem.    If  from  a  fixed  point  without  a  circle  a 
secant  is  drawn,  the  product  of  the  entire  secant  and  the 
external   segment   is   constant   in  whatever  direction  the 
secant  is  drawn. 

166.  Theorem.   If  from  a  point  without  a  circle  a  tan- 
gent and  a  secant  are  drawn,  the  tangent  is  a  mean  propor- 
tional between  the  entire  secant  and  the  external  segment. 

PROBLEMS. 

167.  To  divide  a  given  line  into  parts  proportional  to 
any  number  of  given  lines. 

168.  To  find  the  fourth  proportional  to  three  given  lines. 

169.  To  find  the  third  proportional  to  two  given  lines. 

170.  To  find  the  mean  proportional  between  two  given 
lines. 


SYLLABUS.  XX111 


171.  To  divide  one  side  of  a  triangle  into  two  parts  pro- 
portional to  the  other  two  sides. 

172.  To  divide  a  given  line  in  extreme  and  mean  ratio. 

173.  To  construct  upon  a  given  line  a  triangle  similar 
to  a  given  triangle. 

174.  To  construct  upon  a  given  line  a  polygon  similar 
to  a  given  polygon. 


BOOK  IV. 

EQUIVALENT  FIGURES. 


AREAS. 

175.  Definitions.    Equivalent   figures,  area  of  a  figure, 
units  of  area,  transformation  of  a  figure. 

176.  Theorem.  Two  rectangles  having  equal  bases  are  to 
each  .other  as  their  altitudes ;   and  two  rectangles  having 
equal  altitudes  are  to  each-  other  as  their  bases. 

177.  Theorem.    Any  two  rectangles  are  to  each  other  as 
the  products  of  their  bases  and  altitudes. 

178.  Theorem.   Area  of  a  rectangle  =  base  X  altitude. 

179.  Theorem.    Area  of  a  square  =  square  of  one  side. 

180.  Theorem.    Area  of  a  parallelogram  =  base  X  alti- 
tude. 

181.  Theorem.    Two  parallelograms  having  equal  bases 
are  to  each  other  as  their  altitudes ;  and  two  parallelograms 
having  equal  altitudes  are  to  each  other  as  their  bases. 

182.  Theorem.    Area  of  a  triangle  =  \  base  X  altitude. 

183.  Theorem.    Two  triangles  having  equal  bases  are  to 
each  other  as  their  altitudes ;    and  two  triangles  having 
equal  altitudes  are  to  each  other  as  their  bases. 


XXIV  GEOMETRY. 


184.  TJieorcm.    Area  of  a  trapezoid  =  £  sum  of  the  bases 
X  altitude. 

185.  Theorem.    Area  of  a  trapezoid  =  median  X  altitude. 

186.  Theorem.    Area  of  a  polygon  circumscribed  about 
a  circle  =  £  perimeter  X  radius  of  the  circle. 

COMPARISON  OF  AREAS. 

187.  Theorem.    The   areas   of  two  triangles  having  an 
angle  of  one  equal  to  an  angle  of  the  other  are  to  each  other 
as  the  rectangles  of  the  sides  including  the  equal  angles. 

188.  Theorem.    Similar  triangles  are  to  each  other  as  the 
squares  upon  the  homologous  sides. 

189.  T}ieorem.    Similar  polygons  are  to  each  other  as  the 
squares  upon  the  homologous  sides  or  homologous  diagonals. 

190.  Theorem.    In  a  right  triangle  the  square  upon  the 
hypotenuse  is  equivalent  to  the  sum  of  the  squares  upon 
the  two  legs. 

191.  Theorem.   In  a   triangle  the  square  upon  a  side 
opposite  an  acute  angle  is  equivalent  to  the  sum  of  the 
squares  upon  the  other  two  sides,  diminished  by  twice  the 
rectangle  of  one  of  these  sides  and  the  projection  of  the  other 
upon  it. 

192.  Theorem.    In  an  obtuse  triangle  the  square  upon 
the  side  opposite  the  obtuse  angle  is  equivalent  to  the  sum  of 
the  squares  upon  the  other  two  sides,  increased  by  twice 
the  rectangle  of  one  of  these  sides  and  the  projection  of  the 
other  upon  it. 

PROBLEMS. 

193.  To  find  the  area  of  a  given  polygon.   • 

194.  To  construct  a  square  equivalent  to  the  sum  of  two 
or  more  given  squares. 

195.  To  construct  a  square  equivalent  to  the  difference 
of  two  given  squares. 


SYLLABUS.  XXV 


196.  To  construct  a   polygon   similar  to   two   or  more 
given  similar  polygons  and  equivalent  to  their  sum. 

197.  To  construct  a  polygon  similar  to  two  given  similar 
polygons  and  equivalent  to  their  difference. 

198.  To  construct  a  square  which  shall  have  a  given 
ratio  to  a  given  square. 

199.  To  construct  a  polygon  similar  to  a  given  polygon, 
and  having  a  given  ratio  to  it. 

200.  To  construct  a  triangle  equivalent  to  a  given  poly- 
gon. 

201.  To  construct  a  square  equivalent  to  a  given  paral- 
lelogram. 

202.  To  construct  a  square  equivalent  to  a  given  triangle. 

203.  To  construct  a  square  equivalent  to  any  given  poly- 
gon. 

204.  To  construct  a  polygon  similar  to  a  given  polygon, 
and  equivalent  to  another  given  polygon. 


BOOK   V. 

REGULAR    FIGURES. 


REGULAR  POLYGONS. 

205.  Definitions.    Regular  polygons,  inscribed  and  cir- 
cumscribed regular  polygons. 

206.  Theorem.    A  circle   may   be   circumscribed   about 
every  regular  polygon  and  also  inscribed  in  every  regular 
polygon. 

207.  Definitions.  .Centre  of  a  regular  polygon,  radius, 
apothem,  angle  at  the  centre. 


XXVI  GEOMETRY. 


208.  Theorem.    Each  angle  at  the  centre  of  a  regular 

t  M  36°° 

polygon  of  n  sides  — 

209.  Theorem.    Each  angle  of  a  regular  polygon  is  the 
supplement  of  the  angle  at  the  centre. 

210.  Theorem.   If  a  circumference  is  divided  into  equal 
parts,  the  chords  joining  the  consecutive  points  of  division 
form  an  inscribed  regular  polygon  ;  and  the  tangents  drawn 
through  these  points  form  a  circumscribed  regular  polygon. 

211.  Theorem,.   Two  regular  polygons  of  the  same  num- 
ber of  sides  are  similar. 

212.  Theorem.    The  perimeters  of  two  regular  polygons 
of  the  same  number  of  sides  have  the  same  ratio  as  their 
radii,  or  as  their  apothems. 

213.  Theorem.    Area  of  a  regular  polygon  =  \  perimeter 
X  apothem. 

CIRCUMFERENCE  AND  AREA  OF  THE  CIRCLE. 

The  letter  r  is  used  to  denote  the  radius  of  a  circle. 

214.  Definitions.    Similar  arcs,  sectors,  and  segments. 

215.  Theorem.    The  limit  of  the  perimeter  of  a  circum- 
scribed regular  polygon,  or  of  an  inscribed  regular  polygon, 
when  the  number  of  sides  is  indefinitely  increased,  is  the 
circumference  of  the  circle. 

216.  Theorem.    Two  circumferences  have  the  same  ratio 
as  their  radii,  or  their  diameters. 

217.  Theorem.    The  ratio  of  the  circumference  of  a  circle 
to  its  diameter  is  constant  for  all  circles.     This  ratio  is 
denoted  by  the  Greek  letter  TT. 

218.  Theorem.    Circumference  of  a  circle  =  2 irr. 

219.  Theorem.    Similar  arcs  are  as  their  radii. 

220.  Theorem.    Arc  of  n°  =  -£-  X 

360 

221.  Theorem.    Area  of  a  circle  = 


SYLLABUS.  XXvii 


222.  Theorem.    Two  circles  are  to   each    other'  as   the 
squares  of  their  radii,  or  of  their  diameters. 

223.  Theorem.   Similar  sectors  are  to  each  other  as  the 
squares  of  their  radii. 

224.  Theorem.    Sector  of  n°  =  -^-  X  TTT\ 

360 

225.  Theorem.    Similar  segments  are  to  each  other  as 
the  squares  of  their  radii. 

PROBLEMS. 

226.  To  inscribe  a  square  in  a  given  circle. 

227.  To  inscribe  a  regular  hexagon  and  an  equilateral 
triangle  in  a  given  circle. 

228.  To  inscribe  a  regular  decagon  and  a  regular  pen- 
tagon in  a  given  circle. 

229.  To  inscribe  in  a  circle  a  regular  polygon  of  15  sides. 

230.  To  inscribe  in  a  given  circle  a  regular  polygon 
having  double  the  number  of  sides  of  a  given  inscribed  reg- 
ular polygon. 

231.  Scholium.   Nos.  226-230,  enable  us  to  inscribe  in  a 
circle  regular  polygons,  the  number  of  whose  sides  is  repre- 
sented by  one  of  the  terms  of  the  following  four  series  : 

3,    6,12  .....    3x2n;  4,    8,16  .....    4x2w; 

5,  10,  20  .....    5  X  2n  ;          15,  30,  60  .....  15  X  2W. 

232.  To  circumscribe  about  a   given  circle  a   regular 
polygon  similar  to  a  given  inscribed  regular  polygon. 

233.  To  construct  a  regular  polygon,  given  one  side  and 
the  number  of  sides. 

234.  Given  the  side  a  of  a  regular  polygon  inscribed  in  a 
circle,  to  compute  the  side  b  of  an  inscribed  regular  poly- 
gon having  double  the  number  of  sides. 


235.    To  compute  approximately  the  value  of  TT. 


EXERCISE  MANUAL  IJNT  GEOMETRY. 


CHAPTER   I. 

THE  STRAIGHT  LINE. 


§  1.    QUESTIONS  AND  NUMERICAL  EXERCISES. 

1,  An  angle  is  equal  to  28°  53'  36" ;  find  an  angle  six 
times  as  large.      '^3  °  %/. '  3 6  * 

2,  What   angle   do   the   hands   of  a  clock  make  at  5 
o'clock  ?    /S*Q  * 

3,  What  angle  is  described  by  the  minute-hand  of  a  clock 
in  15  minutes  ?    What  angle  is  described  by  the  hour-hand 
in  the  same  time  ?  J?$o' 

4,  Find  the  values  of  two  adjacent  supplementary  an- 
gles, if  one  is  9  times  as  large  as  the  other.    .'  £&* 

5,  Find  the  supplement  of  32°,  and  the  complement  of 
88°  12' 16".         *>*>*" 

6,  What  is  the  supplement  of  the  complement  of  40°  ? 

the  complement  of  the  supplement  of  91°  1'  1"  ? 
'  /   /  / 

7,  Around  a  point  as  common  vertex  are  arranged  8  an-     . 

gles,  each  angle  being  5°  10'  greater  than  the  one  preced- 
ing it  in  order.     Find  all  the  angles.  #" 

3*    *rF 

8,  Through  the  vertex  of  a  right  angle  a  line  is  drawn 
outside  the  angle.     What  is  the  sum  of  the  two  acute  an-    ;*'    *f 
gles  thereby  formed? 


GEOMETRY. 


9,  One  acute  angle  of  a  right  triangle  is  51°  32' ;  find 
the  other  acute  angle. 

10,  Of  the  angles  of  a  triangle  the  second  is  twice  the  . 
first,  and  the  third  three  times  the  second.     Find  all  the 
angles. 

1L  If  in  a  right  triangle  one  acute  angle  is  A°  greater 
than  the  other,  find  the  value  of  each  angle. 

12,  If  A,  B,  C,  denote  the  angles  of  a  triangle,  find  the 
angles  formed  by  (i.)  the  bisectors  of  A  and  B ;  (ii.)  the 
altitudes  upon  AC  and  BO;  (iii.)  the  bisector  of  A,  and 
the  altitude  upon  BC. 

13,  If  three  angles  of  a  quadrilateral  are  right  angles, 
what  is  the  value  of  the  fourth  angle  ?     ^  o  ° 

14,  What  is  the  sum  of  the  angles  of  an  octagon  ? 

15,  Find  each  angle  of  an  equiangular  decagon.    /  fy£ 

16,  Find  the  sum  of  the  acute  angles  of  a  starred  pen- 
tagon ;  also  the  sum  of  all  the  interior  angles. 

The  starred  pentagon  or  pentagram  is  formed  by  producing  the 
sides  -of  an  equiangular  pentagon,  until  they  meet,  and  then  erasing 
the  sides  of  the  pentagon. 

17,  Find  the  sum  of  the  angles  in  a  polygon  of  20  sides. 

18,  The  sum  of  the  angles  of  a  polygon  is  48  right  an- 
gles ;  find  the  number  of  sides.  7^ 

19,  In  what   polygon   is   the   sum  of  the  angles   three 
times  as  great  as  in  a  pentagon  ?     \\ 

20,  Make  a  quadrilateral  having  the  greatest  possible 
number- of  obtuse  angles.     «^ 

2L  Make  a  hexagon  having  the  greatest  possible  num- 
ber of  reentrant  angles.  J^ 


THE   STRAIGHT    LINE. 


22,  How  many  diagonals  can  be  drawn  from  one  vertex 
of  a  decagon  ?/  How  many  from  all  the  vertices? 

23,  How  many  different  diagonals  can  be  drawn  in  a 

polygon  of  20  sides  ?  ^C"*-^) 

•»- 

24,  If  one  of  the  base  angles  of  an  isosceles  triangle  =  25°, 
find  the  angle  at  the  vertex.     /?J 

25,  What  is  the  value  of  each  acute  angle  in  an  isosceles 
right  triangle  ?     tyf0 

26,  Find  the  angle  formed  by  the  bisectors  of  two  an- 
gles in  an  equilateral  triangle. 


27.  Find  the  angles  of  an  isosceles  triangle  if  a  base 
angle  is  double  the  vertex  angle.    $£  *  j  ^  * 

28.  The  bisector  of  a  base  angle  in  an  isosceles  triangle 
makes,  with  the  opposite  leg,  the  angle  52°  15'  ;  find  all  the 
angles  of  the  triangle    faff 

29.  If  the  angle  at  the  vertex  of  an  isosceles  triangle  is 
•A°t  find  the  angle  formed  by  the  bisectors  of  the  base  an- 
gles.     ;  ';  0.0; 

30.  -Find  the  angles  of  an  isosceles  triangle  if  the  alti- 
tude is  equal  to  half  the  base.    vfjj".  H5.  *|0  - 

31.  In  an  isosceles  triangle  having  a  given  base,  be- 
tween what  limits  must  the  value  of  a  leg  lie?   >^  ,^ 

32.  If  two  equal  right  triangles,  each  having  the  acute 
angles  30°  and  60°,  are  placed  with  their  longer  legs  coin- 
ciding, and  the  other  legs  extending  in  opposite  directions, 
what  kind  of  a  triangle  will  be  formed  ? 


33,    In  a  right  triangle  one  leg  is  equal  to  half  the  hy- 
potenuse ;  find  the  acute  angles.     (See  Ex.  32.)     ^0°  ..^* 


GEOMETRY. 


34,  If  the  acute  angles  of  a  right  triangle  are  30°  and 
60°,  what  angle  is  formed  by  the  bisectors  of  the  acute 
angles?     /$& 

35,  Make  a  triangle,  and  then  draw  the  three  altitudes. 
If  two  of  the  altitudes  lie  without  the  triangle,  what  must 
be  true  of  one  of  the  angles ?    ,1  w,.       •  ;. 

36,  By  drawing  an  altitude  of  a  triangle  (and,  if  neces- 
sary, producing  the  base)  two  right  triangles  are  formed ; 
what  relation  exists  between  these  two  triangles  and  the 
original  triangle,  (i.)  when  the  altitude  lies  within  the  orig- 
inal triangle,  (ii.)  when  it  lies  without  ? 

37,  Can  a  triangle  be  made,  having  for  sides  3  feet,  5 
feet,  and  12  feet?  4  feet,  3  feet,  and  7  feet?  2  feet,  7  feet, 
and  3  feet? 

38,  Two  sides  of  a  triangle  are  12  inches  and  15  inches ; 
between  what  limits  must  the  third  side  lie?     i  ^w^ 

39,  If  one  angle  of  a  parallelogram  is  75°,  find  the  other 
angles.    15       {0.j 

40,  The  difference  between  two  adjacent  angles  of  a  par- 
allelogram is  90°  ;  find  all  the  angles. 

41,  Into  what  figures  is  a  rhombus  divided  by  drawing 
one  diagonal?  two  diagonals?      H  wfa ^AovjJL^  XA. 

42,  What  is  the  sum  of  the  two  angles  of  a  trapezoid 
adjacent  to  one  leg?    IT^J^ 

43,  If  one  angle  of  an  isosceles  trapezoid  is  45°,  find  the 
other  angles.  1  &$ 

44,  If  the  bases  of  a  trapezoid  are  3  inches  and  7  inches 
in  length,  what  is  the  length  of  the  median  ?   £  U*  , 

45,  The  legs  of  a  trapezoid  are  perpendicular  to  each 
other,  and  one  angle  of  the  trapezoid  is  60°  ;  find  the  other 
three  angles.    \  5*0  ,  no  ,  fe  °    $0 


THE    STRAIGHT    LINE. 


§  2.   THEOREMS. 

1,  The  bisectors  of  two  supplementary  adjacent  angles 
are  perpendicular  to  each  other. 

2,  The  bisectors  of  two  vertical  angles  form  one  straight 
line. 

3,  The   bisectors   of  two   alternate-interior   angles   are 
parallel. 

4,  The  bisectors  of  the  acute  angles  of  a  right  triangle 
form  an  angle  of  135°.  / 

5,  The  bisector  of  the  angle  at  the  vertex  of  an  isosceles 
triangle  bisects  the  base  and  is  perpendicular  to  the  base. 

6,  The  perpendicular  drawn  from  the  vertex  of  an  isos- 
celes triangle  to  the  base  bisects  the  base  and  also  the  an- 
gle at  the  vertex. 

7,  The  perpendicular  erected  at  the  middle  point  of  the 
base  of  an  isosceles  triangle  passes  through  the  vertex  and 
bisects  the  angle  at  the  vertex. 

8,  The  bisectors  of  the  base  angles  of  an  isosceles  trian- 
gle form  with  the  base  another  isosceles  triangle. 

9,  If  the  angle  at  the  vertex  of  an   isosceles  triangle  is 
36°,  the  bisector  of  a  base  angle  divides  the  triangle  into 
two  isosceles  triangles.     Find  the  lengths  of  all  the  lines  in 
the  figure,  if  a  leg  of  the  given  triangle  is  denoted  by  a  and 
the  base  by  b. 

10,  The  altitudes  upon   the  legs  of  an  isosceles  triangle 
are  equal.     What  follows  as  to  the  three  altitudes  of  an 
equilateral  triangle? 

11,  The  medians  drawn  to  the  legs  of  an  isosceles  trian- 
gle are  equal. 


6  GEOMETRY. 


12,    The  bisectors  of  the  base  angles  of  an  isosceles  trian- 
gle are  equal. 

v  13,  If  the  angle  at  the  vertex  of  an  isosceles  triangle  is 
a  right  angle,  what  relation  exists  between  the  base  and 
the  altitude  ?  Prove. 

14,  The  perpendiculars  dropped  from  the  middle  point 
of  the  base  of  an  isosceles  triangle  to  the  legs  are  equal. 
/  15,  If  a  leg  of  an  isosceles  triangle  is  produced  from  the 
vertex  by  its  own  length,  arid  the  extremity  joined  to  the 
extremity  of  the  base,  the  joining  line  is  perpendicular  to 
the  base. 

16,  If  through  any  point  in  the  base  of  an  isosceles  tri- 
angle parallels  to  the  legs  are  drawn,  a  parallelogram  will 
be  formed,  having  its  perimeter  equal  to  the  sum  of  the 
legs  of  the  triangle. 

17,  The  sum  of  the  perpendiculars  dropped  from  a  point 
in  the  base  of  an  isosceles  triangle  to  the  legs  is  constant 
and  equal  to  the  altitude  upon  a  leg. 

c 


Fig.  2. 

Let  (Fig.  l)  PD  arid  P^be  the  two  perpendiculars,  BF  the  alti- 
tude upon  the  leg  AC.  Draw  PO  _L  BF,  and  prove  that  the  triangles 
PBG  and  PBD  are  equal. 

18,  The  sum  of  the  perpendiculars  dropped  from  any 
point  within  an  equilateral  triangle  to  the  three  sides  is 
equal  to  the  altitude. 


THE   STRAIGHT    LINE.  7 

Dr&w  through  the  given  point  a  parallel  to  one  side  of  the  tri- 
angle ;  this  reduces  the  theorem  to  Ex.  17. 

19.  The  sum  of  the  lines  which  join  a  point  within  a 
triangle  to  the  three  vertices  is  less  than  the  perimeter,  but 
greater  than  half  the  perimeter. 

Let  ABC  (Fig,  2)  be  the  triangle,  0  the  given  point.     Then,  by 
Nos.  60  and  61, 

;    OA  +  OB<AC 
OB+OC  <AB 
OA-Y  OC  <AB 
Add  these  inequalities,  and  divide  by  2. 

20.  The  sum  of  the  diagonals  of  a  quadrilateral  is  less^ 
than  the  sum  and  greater  than  half  the  sum  of  the  sides. 

21.  Each  side  of  a  triangle  is  less  than  half  the  perim- 
eter.  

22.  The  altitude  of  an  equilateral  triangle   is   greater 
than  half  of  one  side.     (No.  6to.) 

23.  Each  altitude  of  a  triangle  is  less  than  half  the  sum 
of  the  adjacent  sides.     ^Kj3S^  (A 

24.  The  sum  of  the  three  altitudes  of  a  triangle  is  less 
than  the  perimeter.     (Ex.  23.) 

25.  Each  median  of  a  triangle  is  less  than  half  the  sum 
of  the  adjacent  sides. 

Let  ABC  be  the  triangle,  AM  the  median ;  produce  AM  to  D, 
making  MD=  AM,  join  CD,  and  show  that  CD  =  AB. 

26.  The  sum  of  the  three  medians  of  a  triangle  is  less 
than  the  perimeter.     (Ex.  25.) 

27.  If  in   a  right  triangle   one  of  the  acute  angles  is 
equal  to  twice  the  other,  the  hypotenuse  is  equal  to  twice 
the  smaller  leg.     What  are  the  values  of  the  acute  angles  ? 

A  ABC,  right  angle  at  C,  BC  the  smaller  leg.     Construct  Z.  CAD 
=  Z  CAB,  and  produce  BC  to  D.  y  X 


'8  GEOMETRY. 


28,  Prove  the  converse  of  the  preceding  theorem. 

29,  The  bisectors  of  the  angles  of  a  triangle  meet  in  a 
point  equidistant  from  the  sides. 

30,  The  perpendiculars  erected  at  the  middle  points  of 
the  sides  of  a  triangle  meet  in  a  point  equidistant  from  the 
three  vertices. 

31,  Lines  drawn  through  the  vertices  of  a  triangle,  par- 
allel respectively  to  the  opposite  sides,  form  another  trian- 
gle, the  sides  of  which  are  bisected  by  the  vertices  of  the 
first  triangle.      Compare   the  triangles  as  to   magnitude. 

32,  The  three  altitudes  of  a  triangle  meet  in  one  point. 

33,  The  centre  of  a  parallelogram  is  the  middle  point  of 
every  line  drawn  through  the  centre  and  limited  by  the 
sides. 

34,  Every  line  through  the  centre  of  a  parallelogram 
divides  its  perimeter  into  two  equal  parts. 

35,  The  line  which  bisects  two  sides  of  a  triangle  is  par- 
allel to  the  third  side,  and  equal  to  half  the  third  side. 

c 


A  B  A  E 

Fig.  3.  Fig.   4. 

Produce  (Fig.  3)  DE,  making  EF=  DE,  and  join  BF.  &  DOE 
and  BEFarQ  equal,  and  ABDF  is  a  parallelogram. 

36,  A  line  parallel  to  the  base  of  a  triangle,  and  bisecting 
one  side,  bisects  the  other  side,  and  the  part  intercepted  be- 
tween the  two  sides  is  equal  to  half  the  base. 

Through  D  (Fig.  3)  draw  a  line  parallel  to  BO. 


THE   STRAIGHT    LINE.  9 

37,  The  perpendiculars  dropped  from  the  middle  points 
of  two  sides  of  a  triangle  to  the  third  side  are  equal.   (Ex.  36.) 

38,  The  lines  which  join  the  middle  points  of  the  sides 
of  a  triangle  divide  the  triangle  into  four  equal  triangles. 

39,  The  middle  points  of  the  sides  of  a  rectangle  are  the 
vertices  of  a  rhombus. 

Draw  the  diagonals,  and  apply  Ex.  35. 

40,  The  middle  points  of  the  sides  of  a  rhombus  are  the 
vertices  of  a  rectangle. 

41,  If  lines  are  drawn  through  the  middle  points  of  the 
sides  of  a  square  taken  in  order,  what  kind  of  a  figure  will 
they  enclose  ?     Prove. 

42,  The  lines  which  join  the  middle  points  of  the  sides 
of  any  quadrilateral  taken  in  order  enclose  a  parallelogram, 
and  the  perimeter  of  the  parallelogram  is  equal  to  the  sum 
of  the  diagonals  of  the  quadrilateral.     What  kind  of  par- 
allelogram will  be  formed  if  the  quadrilateral  is  a  rectangle  ? 
a  rhombus  ?  a  square  ? 

43,  The  three  medians  of  a  triangle  meet  in  a  point 
which  divides  each  median  into  two  parts  such  that  one  is 
equal  to  twice  the  other. 

In  A  ABC  (Fig.  4)  the  medians  are  AD,  CE,  and  BH.  Let  AD 
and  CE  meet  in  0,  and  JPand  G  be  the  middle  points  of  AO  and  CO 
respectively.  DEFO  is  a  parallelogram,  and  CO  =  20E.  Show  in 
a  similar  way  that  BH  must  also  pass  through  0. 

44,  The  lines  drawn  through  two  opposite  vertices  of  a 
parallelogram,  to  the  middle  points  of  the  opposite  sides, 
divide  one  of  the  diagonals  into  three  equal  parts. 

Show  that  these  two  lines  are  parallel,  and  draw  through  one  of 
the  middle  points  a  line  parallel  to  the  diagonal. 

45,  The  middle  point  of  the  hypotenuse  of  a  right  tri- 
angle is  equidistant  from  the  three  vertices. 


10 


GEOMETRY. 


46.  If  the  diagonals  of  a  parallelogram  are  equal,  the 
figure  is  a  rectangle. 

47.  If  the  diagonals  of  a  parallelogram  are  perpendicu- 
lar to   each  other,   the  figure    is  either  a   rhombus  or  a 
square.     When  is  it  a  rhombus?   when  a  square? 

48.  The  bisectors  of  the  angles  of  a  rhomboid  enclose  a 
rectangle. 

49.  The  bisectors  of  the  angles  of  a  quadrilateral  enclose 
another  quadrilateral  the  opposite  angles  of  which  are  sup- 
plementary.    What  kind  of  quadrilateral  will  be  formed  if 
the  given  quadrilateral  is  a  parallelogram  ?  a  rectangle  ? 

50.  Lines  drawn  through  the  vertices  of  a  quadrilateral, 
parallel  to  the   diagonals,  form  a  parallelogram  twice  as 
large  as  the  quadrilateral. 

51.  The  diagonals  of  an  isosceles  trapezoid  are  equal. 

52.  If  the  diagonals  of  a  trapezoid  are  equal,  the  trape- 
zoid is  isosceles. 


Fig.  5. 


Fig.  6 


Draw  (Fig.  5)  CE  and  DF±  CD.  Show  that  A  ADF  and  BCE 
are  equal,  that  &  COD  and  AOB  are  isosceles,  and  that  &  AOO  and 
SOD  are  equal. 

53.  In  an  isosceles  trapezoid  the  lines  joining  the  middle 
points  of  the  sides,  taken  in  order,  form  a  rhombus. 


THE    STRAIGHT    LINE.  11 


54.  The  line  joining  the  middle  points  of  the  bases  of  an 
isosceles  trapezoid  passes  through  the  intersection  of  the 
diagonals  and  also  through  the  intersection  of  the  two  legs 
produced. 

55.  The  median  of  a  trapezoid  passes  through  the  mid- 
dle points  of  the  two  diagonals. 

56.  The  straight  line  joining  the  middle  points  of  the 
diagonals  of  a  trapezoid  is  equal  to  half  the  difference  of 
the  bases. 

57.  Two  trapezoids  are  equal  if  their  sides   taken   in 
order  are  equal,  each  to  each. 

Divide  each  figure  into  a  triangle  and  a  parallelogram,  and  show 
that  the  two  triangles  are  equal,  and  also  the  two  parallelograms. 

58.  In  any  quadrilateral, 

(i.)  the  lines  joining  the  middle  points  of  the  opposite 
sides  mutually  bisect  each  other  ; 

(ii.)  the  intersection  of  these  lines  bisects  the  line  which 
joins  the  middle  points  of  the  diagonals  ; 

(in.)  the  distance  of  this  intersection  from  any  fixed  line 
in  the  plane  of  the  figure  is  the  arithmetical  mean  of  the 
distances  of  the  four  vertices  of  the  quadrilateral  from  the 
same  fixed  line. 

(i.)  Let  MNPQ  (Fig.  6)  be  the  middle  points  of  the  sides  of  the 
quadrilateral  AS  CD.  The  figure  formed  by  joining  MPNQ  in  order 
is  a  parallelogram  (Ex.  42)  ;  therefore  the  diagonals  MN  and  PQ 
mutually  bisect  each  other  at  their  intersection  0. 

(ii.)  Let  E,  Fbe  the  middle  points  of  AD,  BG.  Show  that  MENF 
is  a  parallelogram  (Ex.  35)  ;  then  EF  divides  MN  into  two  equal 
parts.  Therefore  EF  passes  through  the  point  0. 

(iii.)  Let  ClDl  be  any  fixed  line.  Drop  perpendiculars  to  it  from 
the  points  A,  B,  C,  D,  M,  N,  0.  By  No.  71, 


Therefore,       00,  -  M  +  SBi  +  °°i  +  DDi. 


12  GEOMETRY. 


CHAPTER    II. 

THE   CIRCLE 


§  3.     QUESTIONS  AND  NUMERICAL  EXERCISES. 

1,  What  part  of  a  circle  may  be  called  either  a  segment 
or  a  sector  ? 

2,  How  many  chords   having   a   given    length    can    be 
drawn  through  a  point  within  a  circle  ? 

3,  What  are  the  longest  and  the  shortest  chords   that 
can  be  drawn  through  a  given  point  within  a  circle  ? 

4,  Locate  the  middle  points  of  all  chords  parallel  to  a 
given  line. 

5,  Locate  the  middle  points  of  all  chords  which  have  a 
given  length.     (No.  84.) 

6,  How  many  circles  passing  through  two  given  points 
can    be    described?     Locate    their  centres.     Describe    the 
position  of  the  smallest  circle.     What  is  its  radius  ? 

7,  What  figure  is  formed  by  two  tangents  drawn  through 
an  exterior  point,  and  the  radii  drawn  to  the  points    of 
contact?  ^What  relation  exists  between  the  angle  formed 
by  the  tangents  and  the  angle  formed  by  the  radii  ? 

8,  What  is  the  value  of  an  angle  standing  upon  a  chord 
which  subtends  an  arc  of  150°  ?    Also  of  an  angle  inscribed 
in  a  segment  of  200°  ? 

9,  The  difference  of  two  inscribed  angles  is  4°  ;  find  the 
difference  of  the  corresponding  angles  at  the  centre. 


THE    CIRCLE.  13 


10,  Two  tangents  drawn  through  an  exterior  point  make 
an  angle  of  60°.     Find  the  value  of  the  arc  contained  be- 
tween them.  *f  w 

11,  Through  the  extremities  of  an  arc  of  45°  tangents 
are  drawn ;  find  the  angle  made  by  them. 

12,  Find  the  angle  formed  by  tangents  drawn  through  a 
point,  the  distance  of  which  from  the  centre  is  equal  to 
the  diameter  of  the  circle.  foC 

13,  If  the  tangents  drawn  through  an  exterior  point  are 
perpendicular  to  each  other,  what  is  the  value  of  the  arc 
contained  between  them  ?  ^c 

14,  If  the  angle  between  two  tangents  drawn  through 
an  exterior  point  is  60°,  find  the  angle  between  the  chord 
of  contact  and  one  of  the  tangents.  ^,0°     P  ^    j\ 

15,  If  a  triangle  ABC  is  inscribed  in  a  circle,  and  its 
angles  denoted  by  A,  B,   (7,  respectively,  find  the  angle 
which  the  tangent  through  the  vertex  A  makes  with  the 
side  BC.     Under  what  condition  will  this  tangent  be  par- 
allel to  BC? 

16,  A  circle  and  an  angle  are  situated  in  the  same  plane. 
What  is  the  measure  of  the  angle, 

(i.)    if  the  vertex  is  anywhere  within  the  circle  ? 
(ii.)    if  the  vertex  is  in  the  circumference  ?      G. 

(iii.)   if  the  vertex  is  without  the  circle,  and  both  sides 
of  the  angle  cut  the  circle  ?      ^  .  $J 

(iv.)    if  the  vertex  is  withoufHhe  circle,  and  both  the 

sides  touch  the  circle  ?     cv.  -  <*> 

«v/ 

(v.)    if  the  vertex  is  without  the  circle,  and  one  side  cuts 

while  the  other  side  touches  the  circle?  i 


14  GEOMETRY. 


17,  Describe  the  relative  position  of  two  circles  if  the 
line  of  centres, 

(i.)  is  greater  than  the  sum  of  the  radii ; 
(ii.)  is  equal  to  the  sum  of  the  radii ; 
(iii.)  is  less  than  the  sum  but  greater  than  the  difference 
of  the  radii ; 

(iv.)  is  equal  to  the  difference  of  the  radii ; 
(v.)  is  less  than  the  difference  of  the  radii. 
Illustrate  each  case  by  a  figure. 

18,  Distinguish    between   external   contact  and  internal 
contact  of  two  circles.     Which  kind  of  contact  only  is  pos- 
sible in  the  case  of  two  equal  circles  ? 

19,  Two  circles  touch  each  other,  and  their  centres  are 
8  inches  apart.     The  radius  of  one  of  the  circles  is  equal 
to  5  inches.     What  is  the  radius  of  the  other  ?     (Two  solu- 
tions.) 

20,  If  the  radii  of  two  concentric  circles  are  denoted  by 
a  and  b,  respectively,  find   the    radius  of   a   third    circle 
which  shall  touch  both  the  given  circles  and  contain  the 

smaller. 

>^ 

§  4.     THEOREMS. 

1,  The  radius  which  bisects  an  angle  at  the  centre  bisects 
the  corresponding  chord,  and  is  perpendicular  to  it. 

2,  The  radius  which  bisects  a  chord  bisects  also  the  cor- 
responding arc,  and  is  perpendicular  to  the  chord. 

3,  The  perpendicular  erected  at  the  extremity  of  a  radius 
is  a  tangent  to  the  circle. 

4,  The  perpendicular  dropped  from  the  centre  of  a  cir- 
cle to  a  tangent  passes  through  the  point  of  contact. 


THE   CIRCLE.  15 


5,  The   tangents   drawn    through   the    extremities  of  a 
diameter  are  parallel. 

6,  Two  parallel  lines  intercept  upon  the  circumference  of 
a  circle  equal  arcs. 

7,  In  the  same  circle,  or  in  equal   circles,  if  the   sum 
of  two  arcs  is  less  than  the  circumference,  the  greater  arc 
is  subtended  by  the  greater  chord ;  and  conversely. 

8,  If  an   angle,  the  sides  of  which   pass  through  the 
extremities  of  a  chord,  is  equal  to  the  inscribed  angle  on 
the  same  side  of  the  chord,  the  vertex  of  the  angle  lies  in 
the  circumference  of  the  circle. 

Show  that,  if  the  vertex  were  at  a  point  without  the  circle,  or  at  a 
point  within  the  circle,  we  are  led  to  a  conclusion  which  contradicts 
the  hypothesis. 

9,  If  through  any  point  in  the  convex  arc  included  be- 
tween two  tangents  a  third  tangent  is  drawn,  a  triangle 
will  be  formed,  the  perimeter  of  which  is  constant  and 
equal  to  the  sum  of  the  two  tangents.     (No.  88.) 

10,  If  a  circle  is  inscribed  in  a  triangle  ABC,  and  a,  b,  c, 
denote  the  sides  BO,  AC,  AB,  and  p  half  the  perimeter,       /, 
the  segments  made  by  the  points  of  contact  and  adjacent 

to  the  vertices  A,  B,  C,  respectively,  -a^e  equal  to  p—a, 
p  —  b,  and^  —  c. 

11,  The  perimeter  of  an  inscribed  equilateral  triangle  is 
equal  to  half  the  perimeter  of  the  circumscribed  equilateral 
triangle. 

12,  The  diameter  of  the  circle   circumscribed  about  a 
right  triangle  is  equal  to  the  hypotenuse. 

13,  The  radius  of  the  circle  inscribed  in  an  equilateral 
triangle  is  equal  to  one-third  of  the  altitude  of  the  tri- 
angle. 


16 


GEOMETRY. 


16.  In  a  circumscribed   quadrilateral   the   sum   of  two 
opposite  sides  is  equal  to  the  sum  of  the  other  two  sides. 
(No.  88.) 

17,  A  circle  can  be  inscribed  in  a  quadrilateral  if  the  sum 
of  two  opposite  sides  is  equal  to  the  sum  of  the  other  two 
sides. 


Fig.  7. 


The  bisectors  of  two  angles  A  and  B  (Fig.  7)  meet  at  a  point  O 
equidistant  from  the  sides  AC,  AB,  and  BD  (No.  68)  ;  therefore  a 
circle  can  be  described,  with  0  as  centre,  which  will  touch  these  three 
sides.  The  fourth  side  CD  must  either  (i.)  cut  this  circle,  or  (ii.)  lie 
wholly  without  it,  or  (iii.)  touch  it. 

(i.)  Suppose  CD  cuts  the  circle,  and  CE  a  tangent  passing  through 
C.  Then,  by  the  last  theorem, 

AC'+BE  =  AB+CE 

But  by  hypothesis,  AC+  BD  '=  AB  +  CD' 

Therefore, 
or 


BE-  BD1^  CE  -  CD' 
D*E=CE-CD' 
which  is  impossible  by  No.  60. 

(ii.)   Suppose  CD  lies  without  the  circle.     Then,  by  subtracting  the 
first  of  the  above  equalities  from  the  second,  we  have 

BD-BE=CD-CE, 

or  DE=CD-CE, 

which  is  also  impossible  by  No.  60.     Therefore  CD  must  touch  the 
circle  and  coincide  with  CE. 

18,    In   what   kinds  of  parallelograms   can  a   circle   be 
inscribed  ?     Prove. 


THE    CIRCLE.  17 


19.  The    tangents    drawn    through    the    vertices    of  an 
inscribed  rectangle  enclose  a  rhombus. 

20.  The  tangents  drawn  through  the  vertices  of  an  in- 
scribed quadrilateral,  in  which  two  opposite  angles  are  right 
angles,  form,  a  trapezoid. 

If  the  quadrilateral  is  a  deltoid,  prove  that  the  trapezoid 
will  be  isosceles. 

21.  The  diameter  of  the  circle  inscribed  in  a  right  tri- 
angle is  equal  to  the  difference  between  the  sum  of  the  legs 
and  the  hypotenuse. 

Show  first  that  ODCE  (Fig.  8)  is  a  square.  Then,  CD  +  CE  =  the 
diameter  of  the  inscribed  circle. 

Now,  CD  +  CE=AC+BC-  (AD  +  BE]  \ 

=  AC+BC-(AF  +  BF) 

=  AO+BO-AB. 

22.  The  angle  formed  by  two  tangents  is  equal  to  twice 
the  angle  between  the   chord  of  contact  and  the  radius 
drawn  to  a  point  of  contact. 

23.  If  the  tangents  drawn  from  an  exterior  point  to  a 
circle  form  an  angle  of  120°,  the  distance  of  the  point  from 
the  centre  is  equal  to  the  sum  of  the  tangents. 

24.  An  isosceles  trapezoid  is  inscriptible  ;  that  is,  a  circle 
can  be  circumscribed  about  it. 

25.  If  in  a  circle  two  chords  are  drawn,  and  the  middle 
point  of  the  arc  subtended  by  one  chord  is  joined  to  the 
extremities  of  the    other    chord,   the    two   triangles   thus 
formed  are  equiangular,  and  the  quadrilateral  thus  formed 
is  inscriptible. 

(i.)  The  triangles  have  a  common  angle ;  and  their  other  angles 
are  equal  because  they  have  the  same  measure,  (ii.)  The  opposite 
angles  of  the  quadrilateral  have  for  their  measure  a  semi-circumfer- 
ence. 


18  GEOMETRY. 


26,  If  A,  B,  Cy  A',  B',  C'  are  six  points  in  a  circumfer- 
ence, such,  that  AB  is  parallel  to  A'B',  and  A  O  parallel  to 
A'C',  then  BC'  is  parallel  to  B'C. 

27,  The  bisectors  EF  and   GH  (Fig.  9.)  of  the  angles 
formed  by  the  opposite  sides  of  an  inscribed  quadrilateral 
are  perpendicular  to  each  other. 

By  No.  104,  AF-  CM=  FB  -  MD, 

and  AH-  BN=  CH-  DN, 

whence,  FH-  CM-  BN  =  FB  +  CH-  MN, 

or  FH  +  MN=  IfM  +  FN- 

.-.  ZFOH=  Z  HOM(No.  103) ; 
.-. -E'^isJ.to  GH. 
E 


Fig.  9.  Fig.  10. 


28,  If  a  circle  is  circumscribed  about  an  equilateral  tri- 
angle, and  any  point  in  the  circumference  is  joined  to  the 
three  vertices,  the  greatest  of  the  joining  lines  is  equal  to 
the  sum  of  the  other  two  lines. 

Let  ABC  (Fig.  10)  be  the  triangle,  P  the  point.  Upon  AP  take 
PQ  =  PB.  Then  Z  QBP=  Z  QPB.  But  QPB  =  60°,  (No.  99) ; 
whence  A  PQB  is  equiangular,  and  therefore  equilateral.  Also, 
Z.  AQB=  120°,  (No.  32),  and  Z  BPC=  120°,  (No.  99).  In  tlne&AQB 
and  BPO,  BQ  =  BP,  Z  AQB  =  Z  BPC,  Z  QAB  =  Z  BOP,  (No.  99). 
/.  A  A QB  =  A  BPC,  and  A Q  -  CP.  Hence,  PQ  +  QA  =  PB  +  PC. 


THE   CIRCLE.  19 


29.  Through  a  point  A  without  a  circle  two  secants  are 
drawn :    one,   AOB,  passing  through  the  centre   0 ;    the 
other,  A  CD,  so  that  the  part  AC  without  the    circle  is 
equal  to  the  radius.     Prove  that  the  angle  OAD  is  equal 
to  one-third  of  the  angle  BOD. 

30.  If  a  circle  is  circumscribed  about  any  triangle,  the 
feet  of  the  perpendiculars  dropped  from  any  point  in  the 
circumference    to   the    sides    of    the   triangle   lie    in    one 
straight  line. 

The  circle  (Fig.  11)  having  APfor  diameter  passes  through  E  and 
D  (No.  106.) ;  .-.  Z  APD  -  Z  AED.  Similarly,  it  follows  that  Z  FPC 
=  Z  FEO.  Hence,  the  A  APC  and  DPF  are  equal ;  .-.  Z  APD  = 
Z  FPC.  Hence,  Z  AED  =  Z  FEC;  .:  DE  and  EF  form  one  straight 
line. 


Fig.  II. 

31.  The  altitudes  AD,  BE,   OF,  of  the  triangle  ABC, 
bisect  the  angles  of  the  triangle  DEF. 

The  steps  of  the  proof  are  as  follows  :  the  A  OAF,  OEF  (Fig.  12) 
are  equal,  because  the  figure.  AFOE  is  inscriptible  ;  the  A  OCD, 
OED  are  equal  because  the  figure  EODC  is  inscriptible ;  and  the 
A  OAF,  OCD  are  equal  because  the  figure  AFDC  is  inscriptible. 
Therefore,  Z  OEF=  Z  OED. 

32.  Upon  the  three  sides  of  a  given  triangle  equilateral 
triangles  are  described  outwardly.     Prove  that  the  three 
lines  which  join  their  vertices  to  the  opposite  vertices  of 
the  given  triangle  (i.)  are  equal ;  (ii.)  cut  each  other  at  an- 
gles of  120°  ;  (iii.)  intersect  in  one  point. 


20  GEOMETRY. 


33,  If  ;i  chord  is  divided  into  three  equal    parts,  and 
radii  are  drawn  through  the  points  of  division,  will  these 
radii    divide    the    corresponding    arc    into    equal   parts? 
Prove. 

34,  Let  A  be  any  point  of  a  diameter  ;  B  the  extremity 
of  a  radius  perpendicular  to  the  diameter ;  P  the  point  in 
which  BA  meets  the  circumference  ;   C  the  point  in  which 
the  tangent  through  P  meets  the  diameter  produced.    Prove 
that  A  C=  PC. 

35,  All  chords  of  a  circle  which  touch  an  interior  concen- 
tric circle  are  equal,  and  are  bisected  at  the  points  of  contact. 


Fig.  13.  Fig.  14. 

36,  Two  circles  cut  each  other  in  the  points  P  and  Q^ 
and  through  P  the  line  APB  is  drawn,  meeting  the  cir- 
cles in  A  and  B.     Prove  that  the  angle  AQB  is  constant, 
whatever  be  the  direction  of  APB. 

It  is  sufficient  to  prove  that  the  two  angles  A  and  B  (Fig.  13) 
remain  constant. 

37,  Two  circles  touch  in  the  point  P,  and  through  this 
point  two  lines  are  drawn  meeting  one  of  the  circles  in  A 
and  B,  the  other  in  C  and  D.     Prove  that  the  chords  AB 
and  CD  are  parallel. 

Draw  a  common  tangent  through  P  ( Fig.  14),  and  compare  the 
angles  thus  formed  with  the  angles  A  and  D,  B  and  C. 

38,  Two  circles  touch  in  the  point  P,  and  APB  is  a  line 
meeting  the  circles  in  A  and  B.     Prove  that  the  tangents 
through  A  and  B  are  parallel. 


THE    CIRCLE.  21 


39,  If  two  circles  touch  internally,  and  the  diameter  of 
the  smaller  is  equal  to  the  radius  of  the  larger,  the  circum- 
ference of  the  smaller  bisects  every  chord  of  the  larger 
which  can  be  drawn  through  the  point  of  contaqt. 

40,  Two  circles  touch  internally  in  the  point  P,  and  AB 
is  a  chord  of  the  larger  circle  touching  the  smaller  in  the 
point  0.     Prove  that  PC  bisects  the  angle  APE. 


§  5.    Loci. 

1.  Definitions.  There  are  an  indefinite  number  of  points 
in  a  plane  so  situated  that  they  satisfy  any  one  geometric 
condition.  Points,  however,  taken  at  random,  will  not  sat- 
isfy it.  The  points  which  satisfy  it  are  always  confined  to 
a  certain  line  or  group  of  lines.  This  line  or  group  of  lines 
is  called  the  Locus  of  the  point  which  satisfies  the  given 
condition. 

Since  a  line  may  be  regarded  as  traced  by  a  moving  point, 
we  may  also  define  a  locus  as  the  line  generated  by  a  point     ? 
moving  so  that  it  always  satisfies  a  given  condition.  ) 

The^only  loci  jvhich  are  considered  in  Elementary  Plane 
Geometry  are  those  which  are  either  straight^lines  or  the 
circumferences  of  circles. 


2,  /Scholium.  In  order  to  prove  corr^lslsly  that  a  certain 
line  is  the  locus  of  a  point  which  satisfies  a  given  condition, 
two  propositions  must  be  established  : 

(i.)  that  every  point  in  the  line  satisfies  the  given  condi- 
tion (the  direct  proposition)  ; 

(ii.)  either  that  every  point  satisfying  the  given  condi- 
tion is  in  the  line  (the  converse  proposition),  or  that  every 
point  not  in  the  line  fails  to  satisfy  the  given  condition 
(the  contrary  proposition). 


22 


GEOMETRY. 


3,  The  locus  of  &  point  which  has  a  given  distance  from 
a  given  point  is  the  circumference  of  a  circle  with  the  given 
point  as  centre  and  the  given  distance  as  radius. 

4,  The  Jocus  of  a  point  equidistant  from  £wo  given  points 
is  the   perpendicular  which   bisects   the    line  "joining   the 
points. 

A  C 


Fig.  15. 


Fig  16. 


5.  The  locus  of  a  point  equidistant  from  two  given  inter- 
secting Alines  consists   of   the  two  bisectors  of  the  angles 
formed  by  the  given  lines. 

Let  AS,  CD  (Fig.  15)  be  the  given  lines;  0  their  intersection; 
X  a  point  in  one  of  the  bisectors  ;  MX  and  NX  the  distances  from 
X  to  the  lines.  The  right  triangles  OMX  and  ONX  are  equal ; 
.-.MX=NX. 

Conversely,  let  X  be  a  point  such  that  its  distances  JOT,  NX  from 
'.he  given  lines  are  equal,  and  prove  that  OX  bisects  the  angle  AOC. 

6.  The  locus  of  a  point  equidistant  from  two  given  par- 
jxllel  lines  is  a  line  parallel  to  the  given  lines,  and  equidis- 
tant from  them.  x 

7.  The  locus  of  a  point  which  is  at  a  given  distance  from 
a  given  line  consists  of  the  two  parallels  to  the  given  line, 
drawn  at  the  given  distance  from  it,  one  on  each  side  of  it. 


THE    CIRCLE. 


23 


8,  The  locus  of  the  vertex  of  a  right  triangle,  having  a 
given  hyjpjDkejiufie  as  the  base,  is  the  circumference  described 
upon  the  given  hypotenuse  as  diameter. 

9,  The  locus  of  the  vertex  of  a  triangle,  having  a  given ' 
base  and  a  given  angle  at  the  vertex,  is  the  arc  which 
forms  with  the  base  a  segment  capable  of  containing  the 
given  angle. 

10,  The  locus  of  the  centres  of  the  circles  inscribed  in 
triangles,  having  a  given  base  and  a  given  angle  at  the  ver- 
tex, is  the  arc  which  forms  with  the  base  a  segment  capa- 
ble of  containing  an  angle  equal  to  90°  plus  half  the  given 
angle  at  the  vertex. 

The  bisectors  of  the  angles  CAB  and  CBA,  C^AB  and  C^BA,  etc., 
(Fig.  16),  determine  by  their  intersections  the  centres  0,  Olt  etc.,  of  the 
inscribed  circles.  Then  show  that  the  A  AOB,  AO^  are  constant, 
and  each  equal  to  90°  +  \AGB. 


Fig.  17. 

11.  The  locus  of  the  intersections  of  the  altitudes  of  tri- 
angles, having  a  given  base  and  a  given  angle  at  the  ver- 
tex, is  the  arc  forming  with  the  base  a  segment  capable  of 
containing  an  angle  equal  to  the  supplement  of  the  given 
angle  at  the  vertex. 

Let  ABC  (Fig.  17)  be  one  of  the  triangles,  0  the  intersection  of  the 
altitudes.  Show  that  in  the  quadrilateral  EOFC  the  Z  EOF=  180°  - 
Z  AGB.  Therefore,  the  Z  AOB  =  180°  -  Z  ACS,  and  ia  constant. 


24  GEOMETRY. 

12,  The  locus  of  the  jmiddle  points  of  all  chords  in  a 
given  circle,  which  have  a  given  length,  is  the  concentric 
circumference  having  for  radius  the  distance  from  one  of  the 
chords  to  the  centre. 

13,  In  a  given  circle  having  0  as  centre,  the  locus  of  the 
middle  points  of  chords  which  pass  through  a  given  point 
P  is  the  arc  of  the  circle  having  OP  for  diameter,  and  con- 
tained within  the  given  circle. 

Three  cases  are  to  Le  considered :  (i.)  P  in  the  given  circum- 
ference ;  (ii.)  P  within  the  given  circle  ;  (iii.)  P  without  the  given 
circle. 

14,  The  locus  of  the  extremities  of  tangents  to  a  given 
circle,  which  have  a  given  length^is  the  circumference  of  a 

ickJi  concentric  circle,  the  radius  of  which  is  the  distance  from 
k  ^»   the  centre  to  the  extremity  of  one  of  the  tangents. 

15,  The  locus  of  points,  from  which  tangents  drawn  to  a 
given  circle  form  a  given  angle,  is  the  circumference  of  a 
concentric  circle  having  for  radius  the  distance  from  the 
centre  to  the  intersection  of  any  two  tangents  which  form 
the  given  angle. 

Construct  an  angle  at  the  centre  equal  to  the  supplement  of  the 
given  angle,  and  draw  tangents  through  the  extremities  of  the  radii 
which  form  its  sides  ;  these  tangents  meet  in  one  point  of  the  re- 
quired locus. 

16,  Find  the  locus  of  a  point  which  is  at  a  given  dis- 
tance from  a  given  circumference. 

17,  Find  the  locus  of  the  vertices  of  triangles  having  a 
given  base  and  a  given  altitude. 

18,  Find  the  locus  of  the  middle  points  of  lines  drawn 
from  a  given  point  P  to  a  given  line  L. 

19,  Find  the  locus  of  the  extremities  of  lines  meeting  in 
a  given  point  P  and  bisected  by  a  given  line  L. 

r\ 


THE    CIRCLE. 


25 


20.  Find  the  locus  of  points  the  sum  of  whose  distances 
from  two  given  parallel  lines  is  equal  to  a  given  length. 

Let  d  =  distance  between  the  given  lines,  I  =  the  given  length'. 
If  l>  d,  the  locus  consists  of  the  two  lines  parallel  to  the  given  lines, 
and  lying  without  them,  each  at  the  distance  \  (I  —  d)  from  the  nearer 
line.  lfl  =  d,  every  point  between  the  two  lines  satisfies  the  condi- 
tion. If  I  <  d,  there  is  no  locus. 

21.  Find  the  locus  of  points  the  difference  of  whose  dis- 
tances from  two  given  parallel  lines  is  equal  to  a  given 
length.          }  £  '• 

22.  Find  the  locus  of  points  the  sum  of  whose  distances 
from  two  given  intersecting  lines  is  equal  to  a  given  length. 


Fig.  18. 


Let  the  given  lines  AC,  BD,  meet  in  O  (Fig.  18).  Draw  MN\\  to 
BD,  and  at  a  distance  from  it  equal  to  the  given  length,  and  let  MN 
meet  AC  in  A.  With  O  as  centre,  and  OA  as  radius,  cut  the  given 
lines  in  B,  C,  and  D.  Then  show  that  the  figure  A  BCD  is  a  rect- 
angle, and  that  its  sides  form  the  required  locus. 

23,  Find  the  locus  of  points  the  difference  of  whose  dis- 
tances from  two  given  intersecting  lines  is  equal  to  a  given 
length. 

The  locus  consists  of  the  prolongations  of  the  sides  of  the  rectangle 
ABCD  (Fig.  18). 


26  GEOMETRY. 


24,  An  angle  moves  so  that  its  magnitude  remains  con- 
stant and  its  sides  pass  through  two  fixed  points  ;  find  the 
locus  of  the  vertex. 

25,  A  straight  line  moves  so  that  it  remains  parallel  to 
a  given  line,  and  one  end  touches,  a  given  circle  ;  find  the 
locus  of  the  other  end. 

.     26,    A  ladder  rests  with  one  end  against  a  vertical  wall, 
and  the  other  end  upon  a  horizontal  floor.    If  the  ladder  fall  ? 
by  sliding  along  the  floor,  find  the  locus  of  its  middle  point.  •' 

27,  If  ABC  is  an  equilateral  triangle,  find  the  locus  of  >;,,'./  j 
a  point  P  such  that  PA  =  PB  +  PC.     (§  4,  Ex.  24). 

28,  Given  two  points,  P,  Q,  and  a  straight  line  through 
Q.     Find  the  locus  of  the  foot  of  the  perpendicular  from 
P  to  {;he  ^iven  ling  as  the_latter  revolves  around  Q. 

A-*       V      \ 

29,  AB  is  a  fixed  diameter  of  a  circle,  and  the  chord 
Ada  produced  to  M,  so  that  MC=BC.     Find  the  locus 
of  M  as  .4  C  turns  about  A. 


§  6.     PROBLEMS.     GENERAL  REMARKS. 

1.  A  geometric  problem  requires  the  construction  of  a 
point,  line,  or  figure,  which  shall  satisfy  given  conditions. 

Problems  in  Elementary  Geometry  are  restricted  to  those 
which  can  be  solved  with  the  aid  of  ruler  and  compasses. 

/  2,  A  problem  is  said  to  be  determinate,  if  it  has  one,  two, 
or  any  finite  number  of  solutions  :  indeterminate,  if  it  has  an 
infinite  number  of  solutions;  impossible,  if  it  has  no  solu- 
tion ;  and  over-determinate,  if  more  conditions  than  are  neces- 
sary for  the  solution  are  given. 

3,  A  problem  is  often  possible  for  certain  values  of  the 
given  magnitudes  or  certain  positions  of  the  given  points, 
and  impossible  for  other  values  or  other  positions. 


THE    CIRCLE.  27 


4,  Over-determinate  problems  are  impossible,  except  in 
some  cases  for  particular  values  of  the  given  magnitudes,  or 
particular  positions  of  the  given  points. 

To  construct  a  right  triangle,  having  given  the  two  legs,  or  to 
draw  a  tangent  to  a  given  circle  through  a  given  point  without  the 
circle,  are  examples  of  determinate  problems.  To  find  a  point  equi- 
distant from  two  given  points,  or  to  construct  a  parallelogram,  hav- 
ing given  two  adjacent  sides,  are  examples  of  indeterminate  problems. 
The  problem,  "  to  draw  a  tangent  through  a  given  point  within  a 
given  circle,"  is  always  impossible.  The  problem,  "  to  construct  an 
isosceles  triangle,  having  given  the  base  and  one  leg,"  is  impossible  if 
the  given  leg  is  less  than  half  the  given  base.  The  problem,  "  to 
describe  a  circumference  through  four  given  points,"  is  over-determi- 
nate, and  is  impossible  except  when  the  points  are  so  placed  that  the 
opposite  angles  of  the  quadrilateral  formed  by  joining  the  points  are 
supplementary  (No.  106). 


The  systematic  solution  of  a  problem  consists  of  four 
parts :  the  Analysis,  or  course  of  thought  by  which  the  con- 
struction of  the  required  figure  is  discovered ;  the  Construc- 
tion of  the  required  figure;  the  Proof  that  the  required 
figure  satisfies  all  the  given  conditions ;  the  Discussion  in 
which  we  determine  the  number  of  solutions  for  different 
values  of  the  given  magnitudes,  or  different  positions  of 
the  given  points. 

6,  Problems  differ  so  widely  in  their  nature  that  no 
general  rule  can  be  given  by-  means  of  which  the  analysis 
of  any  problem  may  be  effected.  In  all  cases,  however,  the 
first  step  in  the  analysis  of  a  problem  is  to  construct  a  figure 
representing  the  problem  as  already  solved,  in  order  to  ex- 
amine' more  easily  the  relations  of  the  various  parts.  The 
analysis  is  to  be  considered  finished  when  it  has  been  shown 
how  the  required  figure  may  be  constructed  by  means  of 
the  elementary  problems,  or  by  means  of  any  other  prob- 
lems which  have  been  already  solved. 


28 


GEOMETRY. 


7,  In  general,  the  most  instructive  as  well  as  the  most 
difficult  part  of  the  solution  is  the  analysis ;  and  after  a 
few  problems  have  been  completely  and  systematically 
worked  out,  it  is  advisable  to  consider  a  problem  finished 
when  its  analysis  has  been  effected,  and  the  number  of 
solutions  noted,  and  to  devote  the  time  saved  by  omitting 
the  construction  and  proof  to  the  study  of  new  problems. 

§  7.     PROBLEMS.     CONSTRUCTION  OF  POINTS. 

1.  Number  of  given  Conditions.    Two  given    conditions 
are  necessary  to  determine  the  position  of  a  point. 

2,  Notation.    P,  P1;    P2,  etc.,   denote  given  points;  L, 
Jji,  Z2,  etc.,  denote  given  straight  lines. 

Method  of  Analysis.  The  two  conditions  which  the 
point  must  satisfy  are  considered  separately ;  to  each  of 
them  will  correspond  a  certain  locus  for  the  point.  If 
these  loci  are  known,  or  can  be  found,  the  problem  is 
solved ;  for,  since  the  point  is  in  both  loci,  it  must  be  their 
point  of  intersection.  This  is  the  Method  of  Loci. 


A 
Fig.  19. 


3,  To  find  a  point  X  which  shall  be  equidistant  from  P 
and  P1;  and  at  a  given  distance  d  from  Pt. 

Analysis.  (Fig.  19.)  Let  the  point  X  satisfy  the  given  conditions. 
3ince  X  is  equidistant  from  P  and  Pt,  one  locus  in  which  it  must  lie 


THE    CIRCLE.  29 


is  the  perpendicular  bisecting  PPV  And  since  Xis  at  the  distance  d 
from  Pr  another  locus  is  the  circumference  having  P2  as  centre  and 
d  as  radius.  Therefore  X  is  the  intersection  of  these  two  loci. 

Construction.  Join  PP1(  erect  a  perpendicular  at  A,  the  middle 
point  of  PPj,  and  describe  a  circumference  with  P2  as  centre  and  d 
as  radius.  The  points  X,  F,  in  which  this  circumference  cuts  the 
perpendicular,  satisfy  the  given  conditions. 

Proof.  Join  X  to  P,  Plt  and  P2.  P2X  =  d  (const.).  A  P^X  = 
A  PAX  (No.  39),;  .'.P.X^PX. 

Discussion.  The  problem  has  two  solutions,  one  solution,  or  no 
solution,  according  as  the  circumference  cuts  the  perpendicular,  touches 
it,  or  does  not  meet  it  at  all ;  that  is  (if  P2B  is  ±  to  A  Y),  according 
as  d  >  P2£,  d  =  P2P.,  or  d  <P2B. 

In  L  to  find  a  point  JTso  that : 

4,  X  shall  be  at  the  distance  d  from  P. 

5,  X  shall  be  equidistant  from  P  and  Plf 

6,  X  shall  be  at  the  distance  d  from  Lv 

7,  X  shall  be  equidistant  from  two  given  parallel  lines. 
To  find  a  point  JTsuch  that : 

8,  X  shall  be  at  the  distance  a  from  P,  and  the  distance 
b  from  P!. 

9,  X  shall  be  at  the  distance  a  from  P,  and   at   the 
distance  b  from  L. 

10,  ^  shall  be  equidistant  from  two  parallel  lines,  L  and 
Z-i,  and  at  the  distance  a  from  P. 

11,  JT  shall  be  equidistant  from  two  intersecting  lines, 
L  and  Z/lt  and  at  the  distance  a  from  P. 

12,  ^  shall   be    equidistant   from  P  and  P1(  and  also 
equidistant  from  two  parallel  lines,  L  and  L^ . 

13,  X  shall   be    equidistant  from  P  and  P1(  and  also 
equidistant  from  two  intersecting  lines,  L  and  L\. 


30  GEOMETRY. 


14,  X  shall  be  equidistant  from  two  given  parallel  lines, 
and  also  equidistant  from  two  given  intersecting  lines. 

15,  X  shall  be  equidistant  from  three  given  points. 

16,  X  shall  be  equidistant  from  three  given  lines,  two 
of  which  are  parallel.     (Two  solutions.) 

17,  X  shall  be  equidistant  from  three  given  intersecting 
lines.     (Four  solutions.) 

18,  To  find  a  point  X  in  one  side  of  a  given  triangle 
which  shall  be  equidistant  from  the  other  two  sides. 

19,  To  find  a  point  X  equidistant  from  two  adjacent 
sides  of  a  given  quadrilateral,  and  also  equidistant  from 
the  other  two  sides. 

20,  In  a  given  line  to  find  a  point  from  which  the  lines 
drawn  to  two  given  points  will  be  perpendicular  to  each 
other. 

21,  Given  four  points,  Plt  P2,  P3,  and  P4 ;    to  find    a 
point  Jfsuch  that  the  angles  P^XP-i  and  P3JfP4  shall  each 
be  equal  to  90°. 

22,  To  find  a  point  from  which  the  three  sides  of  a  tri- 
angle are  seen  under  equal  angles. 

This  problem  will  be  impossible  if  one  of  the  angles  of  the  triangle 
is  greater  than  120°.     Why  ? 

23,  Two  villages   are    situated   on    opposite  sides  of  a 
river.     Show  by  construction   how  to  locate  a  bridge  so 
that  it  shall  be  equidistant  from  the  two  villages.     Also 
solve  the  same  problem  for  the  case  in  which  the  villages 
are  on  the  same  side  of  the  river. 

24,  A  man  desires  to  build  a  house  so  that  it  shall  be 
two  miles  from  a  certain  railway  station,  and  also  just  half 
way  between    two   straight   railroads  which   meet  at  the 
station.     Find  by  construction  the  proper  location  for  the 
house.     How  many  solutions  are  there  ? 


THE    CIRCLE.  31 


25,  A  straight  railway  passes  two  miles  from  a  certain 
town.     A  place  is  described  as  four  miles  from  the  town 
and  one  mile  from  the  railway.     Find  by  construction  how 
many  places  answer  the  description.  t\  . 

26,  The  positions  A,  JB,  C  of  three  church-spires  being 
marked  on  a  map,  it  is  required  to  locate  on  the  map  the 
position  M  of  a   house  when   the   angles   A  MB,  BMC, 
formed  by  horizontal  lines  drawn  from  the  house  to  the 
three  spires,  have  been  measured  on  the  ground.  s 

§  8.   PROBLEMS.     CONSTRUCTION  OF  CIRCLES. 

1,  Number  of  given  Conditions.    Tivo   given   conditions    \  I 
determine  a  locus  for  the  centre.      Three  given  conditions 
completely  determine  the  position  and  size  of  the  circle. 

2,  Notation.    K,  K\,  K^  etc.,  denote  circles ;   0,  Ob  02, 
etc.,  denote  their  centres;  r,  rb  ?'2,  etc.,  denote  their  radii. 

3,  Method  of  Analysis.    The  method  of  loci  is  always 
applicable ;  and  in  a  few  cases  auxiliary  points  and  lines 
are  required. 

For  the  sake  of  brevity  the  expression  "  in  K"  will  be  used,  with 
the  meaning  "  in  the  circumference  of  K"  ;  also,  in  future,  when 
ambiguity  of  meaning  is  impossible,  instead  of  "  circumference"  the 
word  "  circle  "  will  often  be  substituted. 

To  find  the  locus  of  the  centre  of  a  circle  which  has  a 
given  radius  r,  and  also  : 

4,  Passes  through  P.  8,    Cuts  L  under  the  chord  a. 

5,  Touches  L.  9,    Cuts  Sunder  the  chord  a. 

6,  Touches  7T  externally.    10,    Cuts  J^in  a  diameter. 

7,  Touches  K  internally.    11.    Cuts  K  at  right  angles. 
To  find  the  locus  of  the  centre  of  a  circle  which  : . 

12,    Passes  through  P  and  Px. 


32  GEOMETRY. 


13,  Touches  L  in  the  point  P. 

14,  Touches  K  in  the  point  P. 

15,  Touches  two  intersecting  lines  L  arid  .£,. 

16,  Touches  two* parallel  lines  .L  and  L^. 

17,  Touches  two  concentric  circles  K  and  ^i. 

In  the  last  two  cases  the  two   given  conditions  determine  the 
radius  as  well  as  one  locus  for  the  centre. 

To  construct  a  circle,  with  the  radius  r  and  the  centre  in 
L,  which  shall : 

18,  Pass  through  P.  19,    Touch  K. 

To  construct  a  circle,  with  the  radius  r  and  the  centre  in 
K,  which  shall : 

20,  Pass  through  P.  21,   Touch  L. 

To  construct  a  circle,  with  the  radius  r,  which  shall : 

22,  Pass  through  P  and  P^. 

23,  Pass  through  P  and  touch  L/^ 

24,  Pass  through  P  and  touch  K. 

25,  Pass  through  P  and  cut  K  in  a  diameter. 

26,  Pass  through  P  and  cut  K  at  right  angles. 

27,  Pass  through  P  and  cut  from  L  a  chord  equal  to  a. 

28,  Touch  L  and  Lv. 

29,  Touch  L  and  K. 

30,  Touch  JTand  Jfi. 

31,  Touch  L  and  cut  JT  at  right  angles. 

32,  Touch  L  and  cut  ./Tin  a  diameter. 

33,  Touch  K  and  cut  KI  in  a  diameter. 


THE   CIRCLE.  33 


34,  Cut  from  L  and  L±  chords  equal  to  a  and  a\. 

35,  Cut  .ZTand  K^  in  diameters. 

36,  Cut  both  JTand  KI  at  right  angles. 

37,  Cut  Km  a  diameter  and  K\  at  right  angles. 

To  construct  a  circle  with  its  centre  in  L  which  shall : 

38,  Pass  through  P  and  Plt 

39,  Touch  A  at  P.  40,   Touch  K  at  P. 

41,  Touch  Li  and  Z2. 

42,  Touch  two  concentric  circles,  TTand  Ky. 

To  construct  a  circle,  with  its  centre  in  K,  which  shall : 

43,  Pass  through  P  and  P,. 

44,  Touch  L  and  Lv.  45,    Touch  L  at  P. 

46,  Touch  Kv  at  P. 

47,  Touch  two  concentric  circles  K±  and  JT2. 

To  construct  a  circle  which  shall  satisfy  the  conditions  in 
Ex.  18,  and  also  : 

48,  Pass  through  P  52,    Touch  JT  externally. 

49,  Touch  L2.  53,    Touch  K  internally. 

50,  Have  its  centre  in  L2.       54.    Cut  K  in  a  diameter. 

51,  Have  its  centre  in  K.       55.    Cut  K  at  right  angles. 

To  construct  a  circle  which  shall  satisfy  the  conditions 
in  Ex.  19,  and  also  : 

56.  Pass  through  P.  60.    Touch  K,  externally. 

57.  Touch  L.  61.    Touch  JT2  internally. 

58.  Have  its  centre  in  L.  62.    Cut  K^  in  a  diameter. 

59.  Have  its  centre  in  7T2.  63.    Cut  _K"2  at  right  angles. 


34 


GEOMETRY. 


64,  To  construct  a  circle  which  shall  touch  L  in  P,  and 
pass  through  Pa. 

65,  To  construct  a  circle  which  shall  touch  K  in  P,  and 
pass  through  P,. 

66,  To  construct  a  circle  which  shall  touch  three  given 
intersecting  lines,  L,  .Li,  and  L^. 


Analysis,  by  means  of  g  5,  Ex.  5.  (Fig.  20.)  By  drawing  the 
bisectors  of  the  exterior  angles  of  the  triangle  ABC,  four  solutions  in 
all  are  obtained.  The  circle  having  its  centre  within  the  triangle 
is  the  inscribed  circle.  The  other  circles  are  called  the  escribed  circles. 
If  the  given  lines  meet  in  a  point,  there  is  no  solution. 

67,  To  construct  a  circle  which  shall  touch  L  in  P,  and 
also  touch  LI. 

Analysis.   Produce  L  and  L^  till  they  meet. 

68.  To  construct  a  circle  which  shall  touch  Z,  and  also 
touch  K\n  P. 

Analysis.  Draw  a  tangent  at  P,  and  the  problem  is  reduced  to 
Ex.  67. 


THE    CIRCLE. 


35 


69,    To  construct  a  circle  which  shall  touch  L  in  P,  and 
also  touch  K. 

A 


Fig.  21. 


Analysis.  (Fig.  21.)  Suppose  the  problem  solved,  and  let  X  be 
the  centre  of  the  required  circle.  One  locus  for  X  is  the  perpendicu- 
lar PA  erected  at  P.  If  AP  is  produced  to  B,  so  that  PB  is  equal 
to  the  radius  00  of  the  given  circle,  then  XB  =  X0\  therefore  X 
must  also  lie  in  the  perpendicular  erected  at  the  middle  point  of 
the  line  OB. 

Another  solution  isv  obtained  by  taking  the  point  B'  in  PA  so 
that  PB'  =  OC,  joining  Off,  and  erecting  a  perpendicular  at  the  mid- 
dle point  of  OB'. 

To  construct  a  circle  with  its  centre  in  Z/,  which  shall : 

70,  Cut  L  in  P  and  touch  1^. 

71,  Cut  L  in  P  and  touch  K 

Analysis  of  Ex.  70.  Suppose  the  problem  solved,  X  the  centre 
of  the  required  circle,  Q  the  point  of  contact  with  Lr  Draw  PA  _L 
to  Z1?  and  show  that  Z  XPQ  =  Z  APQ. 

Analysis  of  Ex.  71.   Similar  to  that  of  Ex.  69. 

§  9.    PROBLEMS.    CONSTRUCTION  OF  STRAIGHT  LINES. 

1,  Number  of  given  Conditions.      Two  conditions  are  nec- 
essary to  determine  a  straight  line. 

2,  Notation.    L,  L{,  L^,  etc.,  denote  straight  lines  given 
in  position  ;  a  and  b  denote  given  lengths. 


36 


GEOMETRY. 


3,  Method  of  Analysis.    If   one  point   in  the   required 
line  is  known,  we  should  try  to  determine  another  point 
by  the  method  of  loci.     If  no  point  in  the  line  is  known, 
or  if  the  method  of  loci  cannot  be  directly  applied,  wo 
must  seek  for  auxiliary  points  or  lines  which  have  known 
relations  to   the  required  line,   and  which  can  be   deter- 
mined by  means  of  loci,  or  known  theorems  ;  in  this  way  the 
problem  can  be  reduced  to  problems  already  solved,  or  easily 
solved,  by  means  of  loci  or  known  theorems. 

Through  a  given  point  in  a  given  line  to  draw  a  line 
making  with  the  given  line,  an  angle  of : 

4,  60°.  6,    120°.  8,   45°.  10,      75°. 

5,  30°.  7,    150°.  9,    15°.  11,    135°. 

12,  To  divide  a  right  angle  into  three  equal  parts. 

13,  Given  the  sum  and  the  difference  of  two  lines,  to 
construct  the  lines. 

Analysis.   Let  a  and  b  denote  the  given  sum  and  the  given  differ- 
ence, x  and  y  the  required  lines.     Then  x  =  ^  (a  +  5),  y  =  \  (a  —  b). 

14,  To  erect  a  perpendicular  at  the  end  of  a  given  line 
without  producing  the  line. 


Fig.  22. 

Analysis.  (Fig.  22.)  Let  AB  he  the  given  line,  and  C  any  point 
equidistant  from  A  and  B.  With  C  as  centre,  and  GA  as  radius, 
describe  a  semicircle,  and  produce  AC  to  meet  it  in  D.  Then  D B 
is  -L  to  AB. 


THE    CIRCLE.  3*7 


15,  To  bisect  the   angle  formed  by  two  lines,  without 
producing  them  to  their  point  of  intersection. 

Analysis.  (Fig.  23.)  Draw  any  secant  cutting  the  lines  in  A  and 
B ;  the  four  bisectors  of  the  angles  at  A  and  B  meet  in  two  points  (7, 
D,  which  are  in  the  bisector  required. 

16,  Explain  how  the  distance  between   two  objects   on 
the  ground,  separated  by  an  inaccessible  swamp,  can  be 
found. 

17,  Explain  how  the  distance  from  an  accessible  object 
to  an  inaccessible  object  can  be  found. 

18,  Given  L  II  to  L^  and  a  point  P;  through  P  to  draw 
a  line  cutting  L  in  A,  and  L±  in  JB,  so  that  AB  =  a.    When 
are  there  two  solutions?  one  solution?  no  solution? 

19,  To  draw  through  the  vertex  of  a  given  angle,  and 
without  the  angle,  a  line  which  shall  make  equal  angles 
with  the  sides  of  the  given  angle. 

Given  P,  L,  and  an  angle  with  its  vertex  at  A  ;  to  draw 
a  line  meeting  the  sides  of  the  angle  in  B,  C,  which  shall 
satisfy  the  following  conditions : 

20,  BC=a,  and  BC\\  to  L. 

21,  £C=a,  w&AB  =  AC. 

22,  BO  passes  through  P,  and  AB  =  AC. 

23,  BO  passes  through  P,  and  PB  =  PC. 

24,  PC" passes  through  P,  and  BC=2AC. 

25,  BC  passes  through  P,  and  BC=  BA. 

Analysis  of  Ex.  20.  Suppose  the  problem  solved  ;  let  AB  cut  L 
in.  A  and  a  parallel  to  AB  through  (7 cut  L  in  E ;  then  BCDE  is  a 
parallelogram. 

Analysis  of  Exs.  21,  22.  The  bisector  of  the  angle  is  perpendicu- 
lar to  the  required  line. 

Discuss  Exs.  22-25  for  :  (i.)  P  within  the  angle;  (ii.)  P  without  the 
angle  ;  (iii.)  P  in  one  side  of  the  angle. 


38  GEOMETRY. 


Given  a  triangle  ABC;  to  draw  a  line  meeting  AC  in 
D,  BO  in  E,  which  shall  satisfy  the  following  conditions: 

26,  DE  II  to  AB,  and  Z>E=  BE. 

27,  DE\\  to  AJB,  and  CD  =  BE. 

28,  DE  II  to  AB,  and  DE=  AD  +  BE. 

29,  DE=a,  and  CD=BE. 

30,  DE--=CD=BE. 

Analysis  of  Ex.  27.    Suppose  the  problem  solved,  and  parallels 
to  AC  drawn  through  B  and  E\  this  reduces  the  problem  to  23. 

Analysis  of  Ex.  29.   Draw  a  parallel  to  DE  through  C,  and  a 
parallel  to  AC  through  E. 

C 


Fig.  24. 

Analysis  of  Ex.  30.  Suppose  the  problem  solved,  and  draw  BD 
(Fig.  24).  The  triangles  BDE  and  DCE  are  isosceles ;  whence, 
/.  DEE  =  BDE  =  \  DEC  =  $  DCE,  which  is  known.  This  deter- 
mines the  point  D,  and  then  ^"is  easily  found,  since  DE=  DC. 

Examine  this  problem  for  the  special  cases  when  Z.  ACE  =90°, 
and  when  Z.  ACE  =120°. 

31,  To  draw  a  line  which  shall  be  equidistant  from  two 
given  points.     How  many  solutions  are  there  ? 

32,  Given  P  and  Pl ;  to  draw  a  line  through  P  so  that 
its  distance  from  Pl  shall  be  equal  to  a. 

33,  Given  P  and  PI  ;  to  draw  a  line  so  that  the  dis- 
tances from  P  and  Plt  respectively,  to  the  line,  shall  be  a 
and  I. 


THE   CIRCLE. 


39 


34,  Given  P,  P1;  P2;  through  P  to  draw  a  line  which 
shall  be  equidistant  from  Px  and  P2. 

The  line  passing  through  P,  and  the  middle  point  of  P^,  is  one 
solution  ;  and  the  line  through  P,  parallel  to  P^,  is  another  solution. 

35,  Given  P,  Pb  P2 ;  through  P  to  draw  a  line  so  that 
the  distances  from  P  to  the   feet   of  the  perpendiculars 
dropped  from  Plt  P2,  to  the  line,  shall  be  equal. 


Fig.  25. 


Fig.  26. 


Analysis.  The  line  through  P,  perpendicular  to  P1P2,  is  one  solu- 
tion. Produce  PlP(Fig.  25)  to  Q,  making  PQ-  PPV  and  join  QP2. 
The  line  APE  JL  to  QP2  is  another  solution  ;  for  the  triangles  P£Pl 
and.  PAQ  are  equal.  .-.  PB  =  PA. 

The  point  Q  is  called  the  symmetrical  p.oint  of  Px,  with  respect 
to  P. 

36.  Given    L   and   two    points,    P,    Px ;    to  draw  lines 
through  the  points  which  shall  meet  in  L,  and  make  equal 
angles  with  L. 

Analysis.  The  points  may  lie  both  on  the  same  side  6f  L,  or  one 
on  each  side  of  L.  In  each  case  there  are  two  solutions.  The  line 
PPl  forms  one  solution.  The  other  solution  is  found  by  the  use  of 
symmetrical  points.  Draw  (Fig.  26)  PA  _L  to  L ;  produce  it,  making 
AQ=  AP,  and  join  QPl  meeting  L  in  P.;  PB  and  PrB  are  the 
required  lines.  The  point  Q  is  the  symmetrical  point  of  P,  with 
respect  to  L. 

37.  To  find  the  shortest  path  from  P  to  Pi,  which  shall 
touch  a  given  line  L. 


40 


GEOMETRY. 


If  the  points  are  on  opposite  sides  of  L,  join  them;  if  on  the  same 
side  (Fig.  26),  make  the  same  construction  as  in  the  last  problem ; 
PBPl  is  the  path  required. 

The  proof  consists  in  showing  that  any  other  path  which  touches 
L,  as  PC+  CP1}  is  greater  than  PB  +  BP^. 

38,  Given  P  and  Pl  within  the  angle  BAO\  to  find  the 
shortest  path  from  P  to  Plt  which  shall  touch  both  sides 
of  the  angle.  (Fig.  27.) 


Fig.  28. 


39,  In  a  game  of  billiards  two  balls  have  the  positions 
A,  B  (Fig.  28) ;  to  find  by  construction  the  path  which  the 
ball  at  A  must  pursue  in  order  to  hit  the  ball  at  B,  after 
first  striking  the  four  sides  of  the  table,  assuming  that  the 
ball  makes  equal  angles  with  the  side  of  the  table  before 
and  aftes  impact. 

Analysis.  Suppose  the  problem  solved,  and  ACDEFB  the  required 
path.  DC  produced  must  pass  through  Alt  the  symmetrical  point  of 
A  with  respect  to  the  first  side  of  the  table ;  ED  through  A2,  the  sym- 
metrical point  of  Al  with  respect  to  the  second  side  ;  etc.  Therefore, 
construct  A^  A%,  A3,  A±\  then  join  BA±,  then  FAZ,  etc. 

40,  To  construct  upon  a  given  line  a  segment  capable  of 
containing  an  angle  of  30°.     What  relation  exists  between 
the  length  of  the  line  and  the  radius  of  the  circle  ? 


THE    CIRCLE.  41 


41,  Given  K  and  L ;  to  draw  a  line  touching  K  and 
II  to  L. 

42,  Given  K  and  L;  to  draw  a  line  touching  K  and 
JLto  L. 

43,  To  draw  a  tangent  through  a  given  point  in  a  given 
arc  without  making  use  of  the  centre. 

Given  jSTand  Pin  K\  to  draw  the  following  lines: 

44,  A  chord  through  P  at  the  distance  a  from  0. 

45,  A  diameter  at  the  distance  a  from  P. 

46,  A  chord  equal  to  b  at  the  distance  a  from  P. 

Analysis  for  Exs.  44,  45.  Describe  a  circle  upon  OP  as  diameter, 
and  apply  No.  101. 

Analysis  for  Ex.  46.   Apply  Exs.  3,  12,  of  \  5,  and  Nos.  80,  87. 

Given  P  within  K\  through  P  to  draw  the  following 
lines : 

47,  The  shortest  chord. 

48,  A  chord  having  a  given  length  a. 

Analysis  for  Ex.  48.  Place  in  jfTany  chord  equal  to  «,  and  let  C 
be  its  middle  point ;  the  required  chord  APS  is  tangent  to  the  circle 
having  O  as  centre  and  00  as  radius.  What  are  the  maximum  and 
minimum  values  of  a  for  which  the  problem  is  possible  ? 

Given  P  without  K]  through  P  to  draw  the  following 
lines : 

49,  A  secant  cutting  Km  A  and  B  so  that  AB  —  a. 

50,  A  secant  cutting  Km  A  and  B  so  that  PA  —  AB. 
Analysis  for  Ex.  49.   See  Analysis  for  Ex.  48. 


42 


GEOMETRY. 


Analysis  for  Ex.  50.  Suppose  PAB  (Fig.  29)  is  the  required  line. 
Produce  OA  by  its  own  length  to  C,  and  join  PC.  A  ACP=  A  AOB; 
.'.  P0=  OA ;  whence  C  can  be  determined.  —  Second  Analysis.  Draw 
AD±  to  PB,  meeting  Kin  D.  Then  DP=  DB,  and  DB  is  a  diameter 
of  K\  whence  B  can  be  determined. 

L 


Fig.  30. 


51.  Given  L,  K,  and  K^  exterior  to  one  another ;  to 
place  between  K  and  KI  a  line  parallel  to  L  so  that  it 
shall  have  a  given  length  a. 

Analysis.  Suppose  the  problem  solved,  and  AB  (Fig.  30)  the 
required  line.  Draw  OC  parallel  and  equal  to  AB,  and  join  OA, 
CB.  The  figure  OABO  is  a  parallelogram,  and  CB  =  OA.  From 
this  the  point  B  is  easily  determined. 


Fig.  31. 

52,  Given  K  and  jf^,  intersecting  each  other ;  to  draw 
through  one  of  the  points  of  intersection  a  line  so  that  the 
two  intercepted  chords  shall  be  equal. 

Analysis.  (Fig.  31.)  Let  APE  be  the  line  required,  AP=  PB\ 
C,  D,  and  M,  respectively,  the  middle  points  of  AP,  PB,  and  the  line 
of  centres  00^  Join  0(7,  MP,  O^D.  00  and  O^D  are  perpendicu- 
lar to  AB,  and  parallel  to  each  other  ;  and,  since  CP=  PD,  MP  is 
the  median  of  the  trapezoid  OCDOr  .'.  1/Pis  perpendicular  to  AB. 


THE    CIRCLE.  43 


53,  Given  K  and  P,  PI  exterior  to  K;  through  P  and 
Pl  to  draw  parallels  cutting  K  so  that  the  intercepted 
chords  shall  be  equal. 

54,  Given  two  circles  exterior  to  each  other ;  to  draw  a 
common  .secant  so  that  the  intercepted  chords  shall  have 
given  lengths  a,  b. 

Analysis.  Inscribe  in  the  given  circles  chords  equal  respectively 
to  a  and  b,  and  describe  circles  concentric  with  the  given  circles  and 
touching  a  and  b.  By  $  5,  Ex.  12,  the  problem  is  now  reduced  to 
No.  131. 


A  C  E 

Fig.  32. 

<* 

55,  Given  two  intersecting  circles  ;  to  draw  through  one 
of  the  points  of  intersection  a  common  secant  which  shall 
have  a  given  length. 

56,  Given  two  concentric  circles  K  and  KI ;  through  a 
given  point  P  in  K  to  draw  a  chord  so  that  it  shall  be  tri- 
sected by  K\. 

57,  Given  an  angle,  and  a  point  P  within  the  angle ; 
through  P  to  draw  a  line  which  shall  form,  with  the  sides 
of  the  angle,  a  triangle  having  a  perimeter  equal  to  a  given 
length  a. 

Analysis.  (Fig.  32.)  Let  BPQ  be  the  required  line.  Construct 
the  escribed  circle  touching  SO  in  F,  and  the  sides  AB,  AC  produced 
in  J).  E.  Since  BD  =  BF,  arid  CE  =  OF,  AB  +  BC+  OA  =  AD  +  AE. 
But  AD  =  AE.  .\  AD  =  AE  =\a.  This  determines  the  circle,  and 
a  tangent  to  the  circle  through  P  is  the  line  required. 


44  GEOMETRY. 


§  10.    PROBLEMS.     CONSTRUCTION  OF  TRIANGLES. 

1.  Parts  of  a  Triangle.    The  direct  parts  are  the  three 
sides  and  the  three  angles.     There  are  numerous  indirect 
parts ;  as,  for  example,  the  three  altitudes,  the  three  medi- 
ans, the  three  bisectors,  the  parts  into  which  the  altitudes, 
medians,  and  bisectors  divide  both  the  sides  and  the  an- 
gles, the  radius  of  the  inscribed  circle,  and  the  radius  of  the 
circumscribed  circle. 

2,  The  Number  of  given  Parts.   A  triangle  can  be  con- 
structed when  its  shape  and  its  size  are  determined. 

The  shape  and  size  are  determined,  in  general,  by  three 
given  parts,  provided  one  at  least  of  the  given  parts  is  a 
length.  To  determine  the  right  triangle  and  the  isosceles 
triangle  two  parts  only,  besides  the  name  of  the  triangle, 
are  required  ;  and  to  determine  the  equilateral  triangle  and 
the  isosceles  right  triangle  one  part  only,  besides  the  name 
of  the  triangle,  is  required. 

9  3,  Notation.  A  ABC  is  the  required  triangle,  and  in 
general.  AB  is  taken  as  the  base,  C  the  vertex.  If  a 
choice  is  necessary,  it  is  to  be  understood  that  BC>  AC. 
Also  a,  b,  c,  respectively,  denote  the  sides  BC,  AC,  AB\ 
a,  /?,  y,  respectively,  the  angles  at  A,  B,  C;  h,  ra,  t,  re- 
spectively, the  altitude  CH,  the  median  CM,  the  bisector 
CT;  p  and  q,  respectively,  the  segments  BH  and  AH\ 
u  and  v,  respectively,  the  segments  JBT&nd  AT.  r  is  the 
radius  of  the  circumscribed  circle,  p  the  radius  of  the  in- 
scribed circle.  The  angle  made  by  two  lines,  as  a  and  h, 
is  denoted  thus  :  /.  ah. 

If  A  ABC  is  a  right  triangle,  c  is  the  hypotenuse,  a  and  b 
the  legs,  y  the  right  angle. 

If  A  ABC  is  isosceles,  c  is  the  base,  a  the  angle  at  the 
base,  y  the  angle  at  the  vertex. 

If  A  A  BC  is  equilateral,  a  is  one  side,  h  the  altitude. 


THE    CIRCLE. 


45 


4,  Method  of  Analysis.  In  the  simplest  cases  the  funda- 
mental problems  (Nos.  112-133  of  the  Syllabus)  can  be 
immediately  applied ;  in  a  few  cases  the  loci  in  Exs.  10  and 
11,  §  5,  are  applicable ;  in  most  cases  the  construction  must 
be  found  by  the  aid  of  other  triangles,  which  can  be  more 
easily  constructed  from  the  given  parts,  and  which  are 
called  auxiliary  triangles. 

It  is  convenient  to  subdivide  the  cases  which  will  be 
here  considered  under  the  heads  A,  B,  0,  etc. 

A,  Cases  in  which  the  loci  of  Ex.  10  and  11  of  §  5  can  be 
applied. 


Let  O  (Fig.  33)  be  the  centre  of  the  circle  circumscribed  about  the 
A  ABC.  Draw  OE  JL  to  AB  •  then  OE  bisects  AB  in  D,  and  the  arc 
AS  in  E.  Therefore,  CE  bisects  the  Z  ACS. 

To  construct  a  right  triangle,  given : 

5,  c,  and  the  altitude  upon  c. 

6,  c,  and  one  segment  of  c  made  by  the  altitude. 

7,  The  two  segments  of  c  made  by  the  altitude. 

8,  The  two  segments  of  c  made  by  the  bisector  of  the 
right  angle. 

9,  c,  and  a  line  L  in  which  C  must  lie. 

10,  c,  and  the  distance  from  O  to  a  line  L. 

11,  c,  and  the  distance  from  O  to  a  point  P. 


46 


GEOMETRY. 


To  construct  a  triangle,  given  : 

12,  c,  k,  y.  15,    u,  v,  y.  18,    c,  a,  /.  bm. 

13,  c,  in,  y.  16,    u,  v,  r.  19,    c,  A,  Z  bin. 

14,  p,  q,  y.  17,    u,  h,  y. 

21,  c,  y,  and  the  condition  that 

22,  c,  and  the  altitudes  upon  the  other  two  sides. 

23,  c,  k,  and  the  altitude  upon  a. 

24,  c,  m,  and  the  altitude  upon  a. 

25,  A,  and  the  distances  from  the  foot  of  h  to  the  sides 
a  and  b. 


20,    c,  y,  Z.  bin. 
—  c. 


B,    Auxiliary  triangles  formed  by  drawing  h,  m,  and  t. 


HTM. 

Fig.  34. 


T       M 
Fig.  35. 


By  drawing  the  altitude  h  two  auxiliary  right  triangles  ACH  and 
B CH  are  formed ;  the  required  triangle  ABC  is  equal  to  their  sum 
if  •<9o4f#sr.  34),  and  equal  to  their  difference  if  a>90°  (Fiy.  35). 
In  the  fir*  case,  c  =  p  +  q  ;  in  the  second  case,  c  =  p  —  q. 

The  median  m  divides  A  ABC  into  the  auxiliary  A  ACM,  BCM. 
It  is  also  the  hypotenuse  of  the  auxiliary  rt.  A  CH^f,  in  which  one 
leg  =  h,  and  the  other  leg  =  $  c  —  q  =  j  ( p  -f  q)  —  q  =  5  (p  —  q). 

The  bisector  t  divides  A  ABC  into  the  auxiliary  &  4CT,  5C77; 
and  the  Z  <c  -  4^(7=  j8  +  £7  -  0  -h  9Q°  -  $(«4-  0)  =  90°  -  J(a  -  j8). 
Hence,  if  a  and  £  are  given,  Z  £c  can  be  constructed.  The  bisector  t 
is  also  the  hypotenuse  of  the  auxiliary  rt,  A  CUT,  in, .which  one 
leg  =  A,  and  the  other  leg  =p  —  u. 


THE    CIRCLE. 


47 


To  construct  a  triangle 

,  given  : 

26, 

Rt.  A,  h,p. 

45, 

rt,  C, 

m. 

64, 

rt, 

^,^. 

27, 

Rt.  A,  rt,  h. 

46. 

C,  7)1, 

£ 

65, 

i» 

w,  y. 

28, 

Rt.  A,  A,  a. 

47. 

rt,   6', 

y 

66, 

«, 

w,  ^. 

29. 

Rt.  A,  rt,  p. 

48. 

6>>  /?, 

Z   CT/l. 

67, 

a, 

w,  Z  rf. 

30. 

Isos.  A,  «,  A. 

49. 

c,  a, 

Z  rt?>i. 

68, 

*, 

w,  Z  d. 

31, 

Isos.  A,  c,  A. 

50. 

c,  w: 

,  Z  am. 

69, 

*, 

A,  rt. 

32. 

Isos.  A,  A,  a. 

51. 

rt,  m 

,h. 

70, 

<, 

4,y. 

33. 

Isos.  A,  A,  y. 

52, 

6,  m 

,  A. 

71, 

«, 

h,  p. 

34. 

Isos.  rt.  A,  A. 

53. 

6',  m 

,X 

72. 

<i 

A,  t6. 

35. 

Equilat.  A,  A. 

54. 

p,  m 

,  A. 

73. 

«, 

t,p. 

36. 

A,  rt,  6. 

55. 

a,p, 

m. 

74. 

u, 

f.ft. 

37. 

A,  p,  a. 

56. 

k,  c, 

Z  cm. 

75, 

a, 

A,  w. 

38. 

h,  a,  ft. 

57. 

rt,  A, 

Z  rtm. 

76, 

rt, 

JVtfr 

39, 

rt,  b,  p. 

58, 

A,  m 

,  Z  rtm. 

77, 

A, 

p;«. 

40, 

rt,  J9,  a. 

59. 

p  —  ( 

^,  rt,  A. 

78. 

A, 

u,ft. 

41, 

7i,  a,  y. 

60. 

p-q,  m,  /?. 

79. 

M, 

p,  a.  —  ft. 

42, 

A,  p,  y. 

61. 

fc'JR 

y- 

80. 

<; 

b,  a  —  ft. 

43, 

rt,  p,  £. 

62. 

w,# 

y- 

81, 

«, 

t,  p  —  u. 

44. 

rt,  A,  C. 

63, 

rt,  £, 

y- 

82, 

A, 

v,  p  —  u. 

c 
/v 

<, 

V 

^\^ 

Fig.  36. 


0.    Auxiliary  triangle  formed  by  producing  the  median 
to  the  length  2m. 


48 


GEOMETRY. 


Let  CD  =  2  CM  (Fig.  36),  and  join  AD,  BD-  then  AGED  is  a  par- 
allelogram. (No.  68.)'  In  the  auxiliary  A  CBD,  BD  =  b,  CD  =  2m- 
Z  CBD  =  180°  -  7.  Also,  Z  ABD  =  a,  Z  JL4I>  =  0,  Z  (MZ>  =  a  +  j8. 
If  DE  is  perpendicular  to  .5(7,  and  D.F  perpendicular  to  -4(7,  then 
Z>£"and  Z>.Fare  equal  respectively  to  the  altitudes  upon  a  and  b  of 
the  A 


To  construct  a  triangle,  given  : 

83,  a,  b,  m.  86,  m,  h,  y.  89,  m,  a,  Z  am. 

84,  a,  m,  y.  87.  m,  a,  a.  90,  m,  Z  aw,  Z  67?!. 

85,  tti,  a,  /?.  88,  a,  m,  Z  6w.       91,  a,  Z  am,  Z  &m. 

92,  m,  h,  and  the  altitude  upon  a. 

93,  m,  a,  and  the  altitude  upon  a. 

94,  m,  and  the  altitudes  upon  a  and  6. 

D,  Auxiliary  triangles  employed  when  sums  or  differ- 
ences of  sides  are  given. 

In  A  ABC  (Fig.  37)  let  BO  AC.  With  C  as  centre  and  GA  as 
radius,  cut  AB  in  F,  BCm  E,  and  j5  C  produced  in  D,  and  draw  AD, 
AE,  OF.  In  the  auxiliary  A  ABD,  BD  =  a  +  b,  /.ADB  =  i  7,  Z  5J.D 
=  a  +  £7  =  o  +  £  (180°  -  a  -  0)  =  90°  +  £  (a  -  0).  In  the  auxiliary 


A  ABE,  BE=a-b, 

BAD  -  90°  =  £(«'  -  JB)  ;    also,  CF=  5,  Z  OffA  -  a.      If  A 

rt.  A  (rt.  Z  at  C),  Z  ^D.8  =  45°,  Z  ^Lj££  =  180°  -  45°  -  135°. 


If,  in  a  rt.  A  (Fig.  38),  ABC,  c  +  b,  or  c  —  b  is  a  given  part,  with 
A  as  centre  and  AB  as  radius,  cut  A  C  produced  in  D  and  in  E,  and 
draw  BD,  BE.  In  the  rt.  A  BCD,  CD  =  c  +  b,  Z  £Z>(7=  J  a ;  in  the 
rt.  A  BCE,  CE=  c-b,  Z  BEC=  90°  -  Z  BDE  =  90°  -  £  a. 


THE    CIRCLE. 


49 


If  the  perimeter  a  +  b  +  c  of  a  triangle  is  given,  produce  AB 
(Fig.  39),  making  AD  =  AC,  BE  =  EC,  and  draw  DC  and  EC.  In 
the  auxiliary  A  DOE,  DE  =  a  +  b  +  c,  Z  CDE  =  }  o,  Z  GtfZ)  =  J  0. 
If  ADCE.c&n  be  constructed,  it  determines  the  required  A  ^.J?(7. 
for  these  two  triangles  have  the  vertex  C  in  common,  and  the  points 
A,  B  are  determined  by  erecting  perpendiculars  at  the  middle  points 
of  DC  and  EC. 

D 


Fig.  40. 


If  a  +  b  —  c  is  a  given  part,  produce  SO  (Fig.  40),  making  CD  = 
CA,  and  upon  BD  take  BE=  SA,  and  draw  AD,  AE.  The  A  ADE 
is  an  auxiliary  triangle  in  which  DE=  a  +  b—  c,  Z.ADE=^.DAC=  %y, 


To  construct  a  triangle,  given  : 


95.  Rt.  A,  a+b,  c. 

96.  Rt.  A,  a+b,  ft. 

97.  Rt.  A,  c+b,  a. 

98.  Rt.  A,  c+a,  ft. 

99.  Isos.  A,  a+h,  c. 

100,  Isos.  A,  a+h,  y. 

101,  Isos.  rt.  A,  a  +  c. 

102,  Equilat.  A,  a  +  h. 

103,  a  +  b,  ft,  y. 

104,  a +  5,  h,  a. 

105,  a +  5,  c,  a. 


106.  a  +  h,  c,ft. 

107.  a  +  b,  c,  a  — ft. 

108.  Isos.  A,  a  +  c,  y. 

109.  Rt.  A,  a  -  b,  c. 

110.  Rt.A,  a-b,  a- 

111.  Rt.A,  £-£,«. 

112.  Rt.A,  a  -A,  ft. 

113.  Isos.  A,  a  —  h,  c. 

114.  Isos.  rt.  A,  a  —  h. 

115.  Equilat.  A,  a  —  h. 

116.  a  -  b,  c,  ft. 


50  GEOMETRY. 


117,  a  -  b,  a,  ft.  124,  Isos.  A,  a  +  b  -f  e-,  y. 

118,  a  -  6,  y,  a  —  /?.  125,  a  -f  6  +  c,  a,  /?. 

119,  a  -  5,  ^,  /?.  126.  a  +  5  +  c,  h,  a. 

120,  a  —  b,  h,  a.  127,  a  +  b  +  c,  A,  y. 

121,  a  —  b,c,  a.  128,  a  -f  &  +  c,  q,  a. 

122,  Isos.  A,  a  —  c,  a.  129,  Rt.  A,  a  -f  b  —  c,  a. 

123,  Rt.  A,  a  +  b  -f  c,  a.  130.  a  +  b  —  c,  a,  ft. 

E,  Auxiliary  triangles  useful  in  certain  cases  in  which 
a  —  ft  or  p  —  q  is  a  given  part. 

In  Fig.  37  draw  the  altitude  OH  of  the  &ABQ  and  join  DF,  FE. 
In  the  A  BCF,  CF=  b,BE  =  p-q,Z  BFC=  180°  -  a,  ABGF=  180° 
-0-(180°-a)  =  a-£.  In  the  A  5DJ1,  5Z>  =  a  -f  ft,  Z.  BDF  == 


180°  -  ^  (180°  -  y)  =  90°  +  £  7.      In    the  A  J5JSF,    BE=a-b, 


. 

If,  in  the  £\  ABC,  a  is  obtuse,  J^and  if  will  lie  in  .ZL1  produced, 

and  BF=p  +  q. 

To  construct  a  triangle,  given  : 

131,  a,  5,  a  —  0.  138,  p-q,a,a  —  ft. 

132,  a,  A,  a  -  0.  139.  a  +  b^-q^ft. 

133,  ^,  0,  a       ft.  140.  a  +  Z>,  p  -  q,  a  -  ft. 

134,  p  —  q,  a,  b.  141,  a  —  b,p  —  q,/3. 

135,  p  —  q,  a,  /?.  142,  Rt,  A,  ^>  —  q,  ft. 

136,  p-q,a,  ft.  143,  Rt,  A,p  —  q,a  —  b. 

137,  p  —  q,a,  h.  144,  Rt.  A,p  —  q,a  —  p. 

F,   Auxiliary  triangles  useful  when  w  or  v,  or  both  u  and 
v,  are  among  the  given  parts. 


THE    CIRCLE. 


51 


In  the  A  ABQ  (Fig.  41)  upon  GB  take  CD  =  CA,  and  join  D  to  the 
foot  Tof  the  bisector  OT.  Also,  draw  DE  II  to  CT.  &DTC=AATC. 
.:DT=AT=v.  A\so,^TDE=^CTD  =  ZATC=ZTED.  .:TE  = 
TD=v.  The  A  BDT  has  the  parts  BD  =  a-b.  BT=  u,  DT=  v, 
/.  BDT=  180° -o,  Z  BTD=a-0;  the  A  BED  has  the  parts  BE  = 
a-b,  BF=u-v,  Z  BDE  =  \ y,  Z  BED  =  180° -  0  -  g  7  =  90°  + 

*(•-«. 


Fig.  42. 


To  construct  a  triangle,  given  : 


145.  w,  -y,  a.  149.  w,  v,  a  —  b.  153.  w,  a  —  b,  a  —  /J. 

146.  u,  v,  (3.  150.  u,  v,  a  —  fl.  154.  a  —  b,u—v,y. 

147.  M,  a,  j8.  151.  M,  a  —  b,a.  155.  M  —  v,  /?,  y. 

148.  v,  a,  y.  152.  v,  t,a  —  /3.  156.  w  —  v,  a  —  6,  a—  ^8. 

G.  Cases  involving  the  circumscribed  and  the  inscribed 
circles. 

The  centre  0  (FigA2)  of  the  circle  circumscribed  about  the  A  ABO, 
is  the  intersection  of  the  perpendiculars  erected  at  the  middle  points 
of  its  sides.  When  the  radius  r  of  this  circle  is  given,  and  also  an 
angle  of  the  triangle  is  either  given  or  can  be  easily  found,  one  side 
of  the  triangle  can  be  constructed  by  means  of  No.  100.  In  some 
cases  the  loci  of  Ex.  10  and  Ex.  11.  §  5,  are  applicable.  In  other  cases 
the  following  relations  will  be  found  useful  : 

Draw  the  altitude  CH}  the  bisector  CT,  the  median  CM,  the  diam- 
eter DOME-  this  diameter  bisects  the  arc  AEB  in  E,  therefore  CT 
produced  passes  through  E.  Also,  draw  CF\\  to  AB,  FGr-Lto  AB, 
and  join  BF,  EF.  AACH=&BFG;  therefore  Z  ABF  =  a,  and 


52  GEOMETRY. 


Z  CBF=  a-0.    The  figure  CHGF  is  a  rectangle ;  therefore  CF=  HG 
=  BH-  BG  =  £#-  ^LS=p  -  </.     Also,  Z  HCT=  AGED  =  £  Z  CEF 


AD  B 

Fig.  43. 

The  centre  0  (Fig.  43)  of  the  circle  inscribed  in  the  A  ABC,  is  the 
intersection  of  the  bisectors  AO,  BO,  CO,  of  the  three  angles.  If  per- 
pendiculars OD,  OE,  OF&re  dropped  from  0  to  the  sides,  the  A  ABC 
will  be  divided  into  six  auxiliary  right  triangles,  in  each  of  which  one 
leg  =  p,  and  one  of  the  acute  angles  =  half  of  one  of  the  angles  of  the 
A  ABC. 

In  problems  in  which  p  is  a  given  part,  the  loci  of  Exs.  5,  7,  and 
15,  of  \  5,  are  sometimes  applicable. 

To  construct  a  triangle,  given  r  and  also  : 

157.  Rt.  A,  a.  167.  p,  q.  177,  u,  v. 

158.  Rt.  A,  a  -  /?.  168,  a,  ft.  178,  h,  p-q. 

159.  Isos.A,c.  169,  h,  y.  179,  m,p  —  q. 

160.  Isos.  A,  a.  170,  m,  y.  180,   t,  p  —  q. 

161.  Isos.  A,  y.  171,  a  -f  ft,  a.  181.  c,  a  —  /?. 

162.  Equilat.  A.  172.  a  -  Z>,  a.  182.  /*,  a  -  ft. 

163.  a,  6.  173,  a  +  b,  y.  183,  h,  m. 

164.  c,  a.  174,  a  —  b,y.  184.  A,  I. 

165.  <?,  A.  175.  a  +  b,c.  185.  a-b,p  —  q. 

166.  c,  w.  176.  a  —  ft,  c.  186,  a  +  ft  +  c,  a. 


THE    CIRCLE.  53 


To  construct  a  triangle,  given  p  and  also  : 

187,  Rt.  A,  a.  192,  Isos.  A,  c.  197,  h,  a. 

188,  Rt.A,  a.  193.  Isos.  A,  y.  198,  h,  y. 

189,  Rt.  A,  h.  194.  Isos.  A,  h.  199,  a,  h. 

190,  Rt.  A,  t.  195.  a,  0.  200.  /?,  h. 

191,  Equilat.  A.  196.  c,  a.  201,  rf,  y. 

§  11.   PROBLEMS.     CONSTRUCTION  OF  QUADRILATERALS. 

1,  Number  of  Given  Parts.   The  trapezium  is  determined 
by  five  parts;  the  trapezoid  by  four  parts;  the  rhomboid,  the 
deltoid,  and  the  isosceles  trapezoid  by  three  parts  ;  the  rect- 
angle and  the  rhombus  by  two  parts ;  and  the  square  by  one 
part. 

2,  Notation.    ABCD  is  any  quadrilateral.    AB  =  a,  BC 
=  b,   CD  =  c,  DA  =  d,  AC=f,  BD  =  g.     The  angles  at 
A,  B,  (7,  D,  respectively,  are  a,  /?,  y,  8.     The  diagonals/,  g, 
intersect  in  E,  and  the  /.  AEB  ==  6.     If  a  circle  can  be 
circumscribed,  r  is  the  radius ;  if  a  circle  can  be  inscribed, 
p  is  the  radius. 

If  the  quadrilateral  is  a  trapezoid,  a  and  c  are  the  bases, 
a>  c,  and  h  the  altitude.  If  the  quadrilateral  is  a  parallel- 
ogram, h  is  the  altitude  upon  AB. 

In  the  rectangle,  a  and  b  are  the  sides ;  in  the  rhombus, 
a  is  the  acute  angle,  J3  the  obtuse  angle ;  in  the  rhombus  and 
the  square,  a  is  the  side. 

3,  Method  of  Analysis.   In  general,  problems  of  this  kind 
are  reducible  to  the  construction  of  triangles  by  drawing- 
one  or  both  diagonals.     The  triangles  require  to  be  con- 
structed only  so  far  as  may  be  necessary  to  determine  the 
unknown  vertices  of  the  quadrilateral.     The  position  of  an 
unknown  vertex  can  often  be  found  by  direct  application 
of  the  Method  of  Loci.     In  some  cases,  especially  in  the 


54  GEOMETRY. 


construction  of  trapezoids,  certain  auxiliary  lines  or  figures 

are  required.  In  constructing  chord  and  tangent  quadri- 
laterals, No.  88,  and  Exs.  3  and  6  of  §  4,  are  to  be  kept  in 
mind. 

A,    Construction  of  parallelograms. 
To  construct  a  square,  given  : 

4,  The  side.               6,  The  diagonal.  7,  /+  a. 

5,  The  perimeter.  8,  /—  a. 

To  construct  a  rectangle,  given  : 

9,  a,b.                    13,  a,Zaf.  17,  a,f+b. 

10,  a,/.                      14,  a,  a  +  b.  l&a,f—b. 

11,  a,  0.                     15,  a  +  b,f.  19,  a  +  b,£  of. 

12,  /,  0.                     16,  a  -  bj.  20,  / -  b,  Z  of. 

To  construct  a  rhombus,  given  : 

21.  f,g.                      25,  a,/.  29,  «-f  A,  a. 

22.  a,  A.                   26,  /  a.  30,  /  +  0,  a. 

23.  A,  a.                    27,  «,/+</.  31,  /+  a,  a. 

24.  A,/.                     28,  a,/-  0.  32,  /-  A,  a. 


To  construct 

a  rhomboid,  given  : 

33,  a,  b,  a. 

41,  aj,g. 

49.  a,/+^,  0. 

34.  a,/,  a. 

42,  /,  #,  0. 

50,  «,/-^,  6». 

35.  a,  &,/. 

43,  /  </,  Z  bg. 

51,  a  +  6,/  a. 

36.  a,  b,  A. 

44,  «,  0,  Z  an. 

52.  a-bjy  a. 

37,  a,  0,  A. 

45,  /  0,  a. 

53.  /+«-,  i,  a. 

38.  /,  A,  a. 

46,   r/,  a,  0. 

54.  /-  a,  5,  a. 

39.  &,/,  A. 

47,  a,  A,  0. 

55,  i  +  A,  a,  a. 

40.  A,  0,  0. 

48,  /  A,  Z  fy. 

56,  b  —  A,/,  a. 

THE    CIRCLE. 


55 


Bi   Construction  of  trapezoids. 

The  relations  given  by  the  following  constructions  are  often  useful. 

Let  ABCD  (Fig.  44)  be  any  trapezoid.  Draw  GFand  BH II  to  AD, 
CG  and  AH  II  to  BD.  Join  H  (the  intersection  of  AH  and  BH}  with 
F,C,G.  The  figures  ACOH,  ADBH,  FCBH,  are  parallelograms.  AF 
=  a-c,  AG  =  AB  +  BG  =  a  +  c,  ZACG'=6,  Z  CAH=/.CGH=  180° 
-  0,  Z  C!F.fl"  =  Z  <?£#=  a  -t-  j8,  Z  J?05  =  Z  FHB  =  180°  -  (a  +  j8). 


To  construct 

57,  a,  b,f. 

58,  a,  5,  A. 

59,  a,f,  a. 

To  construct 

66,  a,  5,  a,  /S. 

67,  a,  b,  d,  a. 

68,  a,f,  g,  a. 

69,  a,  b,  cj. 

70,  a,  5,  h,  a. 

71,  a,  5,  p,  0. 

72,  a,  c,  a,  0. 

73,  a,  5,  <?,  d 

74,  «,  6,  c,  a. 

75,  a,  <?,/,#. 


an  isosceles  trapezoid 

60,  /  A,  «. 

61,  b,  hj. 

62,  a,  A,  a. 

a  trapezoid,  given  : 

76,  «,/,  #,  0. 

77,  b,f,  g,  0. 

78,  /  ff,  a,  0. 

79,  a,  <?,/  0. 

80,  a,  Z>,  c,  0. 

81,  5,  rf,  a,  0. 

82,  6,  a,  /?,  0. 

83,  A,  a,  p,  0. 

84,  ft,  d,  A,  0. 

85,  Z>,  A,  a,  0. 


given  : 

63.  a  +  6,/,  a. 

64.  a  —  b,f,  a. 

65.  £  -f  h,  a,  a. 


86.  b,d,a  +  p,  0. 

87.  a  -  c,  a,  0,  0. 

88.  a-c,b,  d,  0. 

89.  a  +  6,  c,  c?,/. 

90.  a  —  6,  c,/,  a. 

91.  a  +  b,f,g,p. 

92.  a  —  5,  A,  a,  (3. 

93.  a  +  i,  c,  d,  p. 

94.  a  +  c,  b,  d,  a. 

95.  a  —  bj,g,  0. 


56  GEOMETRY. 


0,    Construction  of  chord  quadrilaterals. 

In  the  cases  in  which  r  is  not  given,  it  is  easy  to  construct  a  tri- 
angle, such  that  the  circle  circumscribed  about  this  triangle  will  also 
be  circumscribed  about  the  required  quadrilateral. 

To  construct  a  chord  quadrilateral  given  : 

96,  r,  a,  b,  c.  104,  r,  a  +  b,  c,  d.    112,  a,  b,  c,  a. 

97,  r,  a,  cj.  105,  r,  a  —  b,f,  g.    113,  a,  c,f,  J3. 

98,  r,  a,/,  g.  106,  r,  a  +  &,  a,  0.  114,  c,  0,  y,  0. 

99,  r,  a,  y,  B.          107,  r,  a  -  &,/,  (9.    115,  /,  g,  a,  0. 

100,  r,  a,  5,  0.  108,  a,  6,  c,  /?.  116,  a  +  £,  c,/,  0. 

101,  r,  a,/,  0.  109,  a,  5,  0,  0.  117,  a  -  &,/,  0,  0. 

102,  r,f,  g,  0.  110,  a,  ft,/  0.  118,  /+  a,  5,  c,  0. 

103,  r,/,  a,  0.  Ill,  a,/,  a,  ft.  119,  /-  a,  i,  ft,  y. 

D,    Construction  of  tangent  quadrilaterals. 

When  p  is  not  given,  apply  Ex.  3,  §  4.  In  some  cases  one  side 
and  the  adjacent  angles  are  easily  constructed ;  the  bisectors  of  these 
angles  determine  the  centre  of  the  inscribed  circle. 

In  the  case  of  the  tangent  trapezoid,  the  altitude  of  the  trapezoid 
is  equal  to  the  diameter  of  the  inscribed  circle.  The  median  passes 
through  the  centre  of  the  inscribed  circle  ;  the  same  is  true  of  the 
bisector  of  the  angle  formed  by  the  two  legs.  If  p  and  one  of  the 
legs  are  given,  place  between  the  bases  a  line  equal  in  length  to  the 
given  leg,  and  drop  a  perpendicular  to  this  line  from  the  centre  of 
the  inscribed  circle ;  this  perpendicular  will  meet  the  circle  in  the 
point  of  contact  of  the  given  leg. 

To  construct  a  tangent  trapezoid,  given  : 

120,  p,  a,  b.  124,  a,  />,  0.  128,  p,  a  -f  >»,  d. 

121,  p,  b,  a.  125,  a,  a,  ft.  129,  p,a  —  b,  c. 

122,  p,  aj.  126,  a,  b,  c.  130,  a  -  c,  a,  ft. 

123,  p,  a,  ft.  127,  a,  c,  a.  131,  a  +  c,  a,  ft. 


THE    CIRCLE.  57 


To  construct  a  tangent  quadrilateral,  given  : 

132,  p,  ex,  ft,  y.  138.   p,  C,  a,  ft.  144,    a,f,  a,  ft. 

133,  p,a,b,  y.  139,  p,  a,  e,  a  +  0.  145.  a,/,  /?,  y. 

134,  p,  a,/,  a.  140,  a,  b,  d,  ft.  146,  a  +  ft, /,  a,  0. 

135,  p,/  a,  /?.  141,  a,  5,  cj.  147.  /+  a,  5,  ft,  y. 

136,  p,  a,  b,  a.  142,  a,  d,/,  ft.  148,  a  +  6,  c,/,  £. 

137,  p,  a,  c,  a.  143,  a,  b,  a,  /?.  149,  /-  a,  6,  rf,  ft. 

§  12.  MISCELLANEOUS  EXERCISES. 

1.  If  the  altitude  of  an  equilateral  triangle  is  6  inches, 
find  the  radii  of  the  inscribed  and  the  circumscribed  cir- 
cles. 

2.  If  AC,  BC are  the  legs  of  an  isosceles  triangle,  and 
if  the  circles  described  upon  A  C,  BC  as  diameters,  meet  in 
D,  prove  that  CD  bisects  the  angle  AGE. 

3,  In  every  right  triangle  the  line  joining  the  vertex  of 
the   right  angle  to  the  centre  of  the  square  constructed 
upon  the  hypotenuse  bisects  the  right  angle. 

4,  In  a  triangle   ABC  the  exterior  angles  at  A,  B,  are 
bisected:  Prove  that  the  line  which  joins  the  intersection 
of  the  bisectors  to  the  centre  of  the  inscribed  circle  passes 
(if  produced)  through  the  vertex  C. 

5.  How  many  circles  can  be  described  touching  the  sides 
of  a  triangle  or  the  sides  produced  ?     How  many  lines  of 
centres  can  be  drawn  ?     Prove  that  the  circumference  of 
the  circumscribed  circle  passes  through  the  middle  point 
of  every  line  of  centres. 

6,  The  feet  of  the  altitudes  and  of  the  medians  of  a  tri- 
angle lie  in  the  same  circumference. 


58  GEOMETRY. 


7,  To  find  the  locus  of  the  vertex  of  a  triangle,  given 
one  side  and  the  length  of  the  corresponding  median. 

8,  To  construct  a  triangle,  given  the  middle  points  of 
the  sides. 

9,  To  construct  a  triangle,  given  the  three  medians. 

10,  To  construct  a  triangle,  given  h,  m,  t. 

11,  To  construct  a  triangle,  given  the  feet  of  the  three 
altitudes. 

12,  To  inscribe  a  rectangle  in  a  circle,  given  a-\-b. 

13,  To  inscribe  a  rectangle  in  a  circle,  given  a  —  b. 

14,  To  construct  three  equal  circles  about  a  given  circle 
so  that  each  shall  touch  the  other  two,  and  also  the  given 
circle. 

15,  To  construct  three  equal  circles  within  a  given  circle 
so  that  each  shall  touch  two  others,  and  also  the  given 
circle. 

16,  To  construct  four  equal  circles  about  a  given  circle 
so  that  each  shall  touch  two  others,  and  also  the  given 
circle. 

17,  To  construct  four  equal  circles  within  a  given  circle 
so  that  each  shall  touch  two  others,  and   also  the  given 
circle. 

18,  To  construct  three  equal  circles  in  an   equilateral 
triangle  so  that  each  shall  touch  the  other  two,  and  also 
two  sides  of  the  triangle. 

19,  To  construct  four  equal  circles  in  a  square  so  that 
each    shall   touch   two  others,    and  also  two  sides  of  the 
square. 

20,  In  a  given  triangle  to  construct  a  semicircle  having 
its  diameter  on  one  side,  and  touching  the  other  two  sides. 


THE    CIRCLE.  59 


21,  To  inscribe  a  circle  in  a  given  sector. 

22,  To  construct  an  equilateral  triangle  so  that  its  ver- 
tices shall  lie  in  three  given  parallel  lines. 

23,  In  a  square  AEGD  to  construct  an  equilateral  tri- 
angle AEF,  so  that  -Z?and  F  shall  lie  in  the  sides  of  the 
square. 

24,  To  cut  off  the  corners  of  a  square  by  straight  lines 
in  such  a  way  that  a  regular  octagon  shall  be  formed. 

25,  Through  the  vertices  of  a  given  equilateral  triangle 
to  draw  lines  which  shall  form  another  equilateral  triangle 
having  a  given  side. 

26,  In  a  given  square  to  construct  a  square  having  a 
given  side  so  that  its  vertices  shall  lie  in  the  sides  of  the 
given  square. 

27,  To  inscribe  a  square  in  the  part  common   to  two 
equal  intersecting  circles. 

28,  Through  a  given  point  A  in  the  plane  of  a  given 
circle  any  secant  AEG  is  drawn.     At  the  middle  point  M 
of  BC  a  perpendicular  HP  equal  to  MA  is  ejected.     Find 
the  locus  of  the  point  P. 

29,  Given  a  circle  and  two  parallel  secants ;  to  draw  a 
tangent  so  that  the  part  contained  between    the    secants 
shall  be  bisected  at  the  point  of  contact. 

30,  Given  a  circle  and  two  lines,  OA,  OB,  meeting  at 
the  centre  0 ;  to  draw  a  tangent  AB  so  that  the  part  con- 
tained between  the  lines  shall  have  a  given  length. 

Given  P,  PI,  P2 ;  through  P  to  draw  a  line  so  that,  if 
perpendiculars  P\X,  P2  ]Tare  dropped  to  the  line, 

31,  PX+PY=a.  32,   PX-  PY=b. 


60  UKOMETRY. 


CHAPTEE  III. 

SIMILAR    FIGURES. 


§  13.   THEOREMS. 

1,    If  three  lines  divide  two  parallels  into  proportional 
parts,  these  lines  meet  in  one  point. 


Let  AB  and  *CD  meet  in  0  (Fig.  45).    In  the  similar  A  OAC,  OBD, 
AC:  BD  =  CO  :  DO. 

Let  CD  and  EFmeet  in  Q.     In  the  similar  A  QCE,  QDF,  CE:  DF 
=  CQ  :  DQ.     But,  by  the  hypothesis, 

AC:  BD=  CE-.DF. 

Therefore,  CO  :  DO  =  CQ  :  DQ ; 

a  proportion  from  which  it  follows  that  0  and  Q  must  coincide.    For, 
by  the  Theory  of  Proportions, 

CO-DO-.DO=CQ-DQ:  DQ, 
or,  CD -.  DO  =  CD  :  DQ- 

whence,  DO  =  DQ. 

2,    Any  two  altitudes  of  a  triangle  are  inversely  propor- 
tional to  the  corresponding  bases. 


SIMILAR    FIGURES. 


61 


3,  If  the  line  joining  the  middle  points  of  the  bases  of  a 
trapezoid  is  produced,  and  the  two  legs  are  also  produced, 
the  three  lines  will  meet  in  the  same  point. 

4,  If  a  line  drawn  from  one  vertex  of  a  triangle  divides 
the  opposite  side  into  parts  proportional  to  the  adjacent 
sides,  the  line  bisects  the  angle  at  the  vertex, 

This  theorem  is  the  converse  of  No.  143.     Either  a  direct  or  indi- 
rect proof  may  be  given. 

5,  State  and  prove  the  converse  of  No.  144. 

6,  In  a  quadrilateral  A  BCD,  in  which  the  angles  at  B 
and  D  are  right  angles,  perpendiculars  PE,  PF&Ye  dropped 
from  any  point  P,  in  the  diagonal  AC,  to  the  sides  BC, 
AD,  respectively.     Prove  that 


AB      CD 


Fig.  46. 


7,  If  in  a  triangle  ABC  any  length  AD  is  taken  from 
one  side  AC,  and  an  equal  length  BE\§  added  to  the  side 
CB,  the  new  base  DE  is  divided  by  the  base  AB  in  the 
inverse  ratio  of  the  sides  AC  and  BC. 

Draw  (Fig.  46)  DF II  to  AB,  and  apply  No.  139. 

8,  In  a  circle  a  line  EF  is  drawn  perpendicular  to  a 
diameter  AB,  and  meeting  it  in  Gr.    Through  A  any  chord 
AD  is  drawn,  meeting  EFin.  C.     Prove  that  the  product 
AD  X  A  C  is  constant,  whatever  be  the  direction  of  AD. 

Join  BD  (Fig.  47)  and  compare  the  A  ACG,  ADB.     Is  the  theo- 
rem also  true  when  G  lies  outside  the  circle  ? 


62 


GEOMETRY. 


9,  The  squares  of  two  chords,  drawn  through  the  same 
point  in  a  circumference,  have  the  same  ratio  as  their  pro- 
jections  upon    the    diameter    drawn    through   that   point. 
(No.  160.) 

10,  If  two  lines  OA,  OB,  drawn  through  a  point  0,  are 
divided  in  C,  D,  respectively,  so  that  OA  X  00=  OBx  OD, 
a  circle  can  be  described  through  the  points  A,  B,  C,  D. 

Show  that(.%.  48)  the  &  DAO,  CEO  are  similar,  and  the  A  DAO 
CBO  equal.  Therefore,  if  a  segment  be  described  upon  CD  capable 
of  containing  the  Z  DAO,  the  arc  of  this  segment  will  pass  through  B. 

0 


Fig.  48. 


Fig.  49. 


11,  .If  in  a  parallelogram  ABCD  a  secant  DE  is  drawn, 
meeting  the  diagonal  AC  in  F,  the  side  BC in  G,  and  the 
side  AB  produced  in  E,  then  DF2  =  FG  X  FE. 

The  &  AFE,  DFC  (Fig.  49)  are  similar ;  and  also  the  &  AFD, 
CFG. 

12,  The  sum  of  the  squares  of  the  segments  formed  by 
two  perpendicular  chords  is  equal   to  the  square  of  the 
diameter  of  the  circle. 

(Fig.  50.)     Apply  No.  161 ;  also  show  that  EC=  AD. 

13,  If  three  circles  mutually  intersect  one  another,  the 
common  chords  pass  through  the  same  point. 

Let  M,  N,  £  (Fig.  51)  denote  the  circles,  and  let  the  chords  CD, 
EF meet  in  0.  Join  AO,  and  suppose  that  AO  produced  does  not 
pass  through  B,  but  through  Pin  J/and  Q  in  N.  Then  we  have, 


SIMILAR    FIGURES. 


In.fi, 
In  M, 

In  N, 
Whence, 


OCX  OD=OEx  OF. 

OCxOD=OAxOP. 

OExOF=OAxOQ. 

OP*=  OQ, 


a  relation  which  cannot  be  true  unless  P  and  Q  coincide  with  B. 


p 
Fig.  51. 

14,  In  every  triangle  the  intersection  of  the  three  alti- 
tudes, the  intersection  of  the  three  medians,  and  the  inter- 
section of  the  three  perpendiculars  erected  at  the  middle 
points  of  the  sides,  lie  in  a  straight  line ;  and  the  distance 
between  the  first  two  points  is  double  the  distance  between 
the  last  two. 


Let  D,  E,  F(Fig.  52)  be  the  three  points  in  question.  The  line 
MN,  joining  the  middle  points  of  AB,  EG,  is  II  to  AC.  Hence,  the 
&  ADC  and  MNF  are  similar,  and 

NF=  MN^  1 
AD      AC     2 


But 


||=|.    (§2,  Ex.  43.) 


Hence,  the  &ADE&K&  NEF  are  similar,  and  ZAED  =  Z  NEF- 
then  DEF  is  a  straight  line.     Also,  from  the  &  ADE,  NEF, 
EF  ^EN^l 
ED     EA      2 


64  GEOMETRY. 


15,  Two  circles  cut  in  point  P.     Through  P  three  lines 
are  drawn,  meeting  one  of  the  circles  in  A,  B,  C,  the  other 
in  D,  E,  F,  respectively.     Prove  that  the  triangles  ABC, 
DEF  %XQ  similar. 

16,  In  every  triangle  the  product  of  two  sides  is  equal  to 
the  product  of  the  diameter  of  the  circumscribed  circle  and 
the  altitude  upon  the  third  side. 

If  AC,  BC are  taken  as  the  two  sides,  draw  the  diameter  CE,  and 
join  CB. 

17,  In  every  inscribed  quadrilateral  the  product  of  the 
diagonals  is  equal  to  the  sum  of  the  products  of  the  oppo- 
site sides. 

Let  ABGD  be  the  quadrilateral.  In  A  C  take  a  point  ^such  that 
Z  EDC=  Z.  ADB.  &  ADB  and  CDE  are  similar ;  also,  the  &  BCD 
and  ADE.  From  these  triangles  obtain  equations  involving  the 
sides,  and  then  add  them. 

18,  In  every  triangle  the  product  of  two  sides  is  equal 
to  the  square  of  the  bisector  of  the  included  angle  plus  the 
product  of  the  segments  into  which  it  divides  the  third 
side. 

If  AC,  BC  are  the  two  sides,  and  CD  the  bisector,  produce  CD  to 
meet  the  circumscribed  circle  in  E,  and  join  BE.  A  A  CD  and  ECB 
are  similaf ;  also,  see  No.  164. 

19,  In  every  triangle  the  sum  of  the  squares  of  two  sides 
is  equal  to  twice  the  square  of  half  the  third  side  plus  twice 
the  square  of  the  median  drawn  to  the  third  side.  (Nos.  162 
and  163.) 

20,  In  every  triangle  the  difference  of  the  squares  of  two 
sides  is  equal  to  twice  the  product  of  the  third  side  and  the. 
projection  of  the  median  upon  the  third  side. 

21,  In  the  diameter  of  a  circle  two  points  A,  B  are  taken 
equally  distant  from  the  centre,  and  joined  to  a  point  P  in 
the  circumference.     Prove  that  the  sum  AP  -f-  BP  is  con- 
stant for  all  positions  of  P.     (Ex.  19.) 


SIMILAR    FIGURES.  65 


22,  The  sum  of  the  squares  of  the  sides  of  a  parallelo- 
gram is  equal  to  the  sum  of  the  squares  of  the  diagonals 
(Ex.  19.) 

23,  The  sum  of  the  squares  of  the  sides  of  any  quadri- 
lateral is  equal  to  the  sum  of  the  squares  of  the  diagonals 
plus  four  times  the  square  of  the  line  joining  the  middle 
points  of  the  diagonals.     (Ex.  19.) 

24,  The  sum  of  the  squares  of  the  diagonals  of  a  trape- 
zoid  is  equal  to  the  sum  of  the  squares  of  the  legs  plus 
twice  the  product  of  the  bases.     (Ex.  22  and  §  2,  Ex.  56.) 

25,  Two  triangles  are  similar  if  their  sides  are  parallel 
each  to  each.     Which  are  homologous  sides  ? 

26,  Two  triangles  are  similar  if  their  sides  are  perpen- 
dicular each  to  each.     Which  are  homologous  sides  ? 

27,  If  two  similar  triangles  ABC,  Z^.Fhave  their  homol- 
ogous sides  parallel,  the  lines  AD,  BE,  CF  which  join  their 
homologous  vertices  meet  in  the  same  point. 


Fig.  53. 

•  Let  AT)  and  CF  (Fig.  53)  meet  in   0,  BE  and  GFin  Q.     From 
the  similar  triangles  in  the  "figure,  prove  that 
CO  :  FO  =  CQ  :  FQ  ; 

whence,  GO  -  FO  -.  FO  =  CQ  -  FQ  •  FQ. 

Or,  CF  •.  FO  =  CF -.  FQ. 

:.FO=  FQ,  or  Q  coincides  with  0. 


66 


GEOMETRY. 


28,  Two  polygons  are  similar  if  their  sides  are  parallel 
each  to  each. 

29.  If  two  similar  polygons  have  their  homologous  sides 
parallel,  the  lines  joining  their  homologous  vertices  meet  in 
the  same  point. 

Let  two  of  the  lines  AA1,  BB'  meet  in  0,  and  let  AA'  meet  any 
third  line,  as  DD',  in  Q.  Draw  diagonals,  and  prove,  as  in  Ex.  24, 
that  Q  must  coincide  with  0. 

The  ratio  AO  -.  A'O,  BO :  B'O,  etc.,  are  each  equal  to  the  ratio  of 
any  two  homologous  sides. 


Fig.  55. 


The  point  O  is  called  the  centre  of  similitude  of  the  polygons.  If 
the  homologous  sides  are  directed  the  same  way  (Fig.  54),  the  two 
polygons  are  said  to  be  similarly  plated,  and  0  is  called  the  direct 
centre  of  similitude.  If  the  homologous  sides  are  directed  opposite 
ways  (Fig.  55),  the  two  polygons  are  said  to  be  inversely  placed,  and 

O  is  called  the  inverse  centre  of  similitude. 

• 

30,  If  a  point  0  is  joined  to  the  vertices  of  a  polygon 
ABODE,  and  upon  the  lines  OA,  OB,  etc.,  the  lengths 
OA',  OB1,  etc.,  are  laid  off,  so  that  the  ratios  OA' :  OA, 
OB1:  OB,  etc.,  are  equal,  the  polygon  A'B'C'D'E1  is  sim- 
ilar to  the  polygon  ABODE. 


SIMILAR    FIGURES.  67 


31.  If  a  point  0  is  joined  to  the  vertices  of  a  polygon 
ABCDE,  and  through  any  point  A'  in  OA  a  line  parallel 
to  AB  is  drawn,  meeting  OB  in  B\  and  through  B1  a  line 
parallel  to  BC,  meeting  00 in  C",  etc.,  the  polygon  A'B'C' 
D'E'  is  similar  to  the  polygon  ABCDE. 

DEFINITIONS.  Let  AB  be  a  given  line,  ra :  n  a  given  ratio,  P  a 
point  in  AB,  such  that  PA  :  PB  =  m-.n\  then  .4.5  is  said  to  be 
divided  internally  or  externally  in  the  ratio  ra  :  n,  according  as  P  is 
between  A  and  B  or  in  J..Z?  produced. 

If  the  line  joining  the  centres  of  two  circles  is  divided  externally 
and  internally  in  the  ratio  of  their  radii,  the  points  of  division  are 
called  the  direct  and  the  inverse  centres  of  similitude,  respectively,  of 
the  two  circles. 

It  follows  from  these  definitions  that  the  point  of  contact  of  two 
circles  which  touch  externally  is  an  inverse  centre  of  similitude  ;  and 
that  the  point  of  contact  of  two  circles,  one  of  which  touches  the  other 
internally,  is  a  direct  centre  of  similitude. 

32,  The  line  joining  the  extremities  of  two  parallel  radii 
of  two  circles  passes  through  the  direct  centre  of  similitude, 
if  the  radii  have  the  same  direction  ;  and  through  the  in- 
verse centre  if  the  radii  have  opposite  directions. 


Fig.  56. 


Let  r,  r1  (Fig.  56)  denote  the  radii  of  the  circles,  0,  O  their  cen- 
tres ;  and  let  OM  be  II  to  OR,  ON\\  O'T,  RM  and  OO1  meet  in  P. 
OTandOO'ineetin  Q.  Then,  OP:  O>P  =  r  •  r1 ,  and  OQ:  0>Q,  =  r  •  r1 . 
.',  Pis  the  direct,  and  Q  the  inverse,  centre  of  similitude. 


68  GEOMETRY. 


33,  The  two  radii  of  one  circle,  drawn  to  its  points  of 
intersection  with  any  line  passing  through  a  centre  of  simili- 
tude, are  parallel,  respectively,  to  the  two  radii  of  the  other 
circle,  drawn  to  its  intersections  with  the  same  line. 

Proof  indirect  with  the  aid  of  Ex.  32. 

34,  All  secants,  drawn  through  the  direct  centre  of  simil- 
itude P  of  two  circles,  cut  the  circles  in  points  whose  dis- 
tances from  P,  taken  in  order,  form  a  proportion. 

Let  the  line  of  centres  (Fig.  56)  cut  the  circles  in  the  points  A,  B, 
C,  D,  and  let  any  other  secant  through  P  cut  the  circles  in  the  points 
Mt  N,  R,  S. 

For  the  line  of  centres  we  have,  by  definition, 

OP:  0'P=r:  r>  ; 

whence,  OP-  r  :  OP  +  r  =  O'P  -  r»  :  O'P  +  r1 

or,  PA:  PB  =  PC:  PD. 

For  any  other  secant  we  have,  by  similar  triangles, 

PM  :  PR  =  PN  :  PS  =  r  :  r'. 
.-.  PM:  PN  =  PR:  PS. 

35,  Using  the  notation  of  the  last  exercise,  prove  that 
the  product  PN  X  PR  is  constant   for  all  secants,  and 
equal  to  the  product  PB  X  PC. 

Join  MA  and  EC  (Fig.  56),  and  show  that  the  A  PA  M  and  PCR 
are  similar  ;  therefore, 

PA  :  PM  =  PC:  PR. 

But  PA  :  PM=  PN:  PB. 

.'.PN-  PB  =  PC:PR, 

and 


36,  The  common  exterior  tangents  to  two  circles  pass 
through  the  direct  centre  of  similitude,  and  the  common 
interior  tangents  pass  through  the  inverse  centre  of  simili- 
tude. 

What  method  of  drawing  the  common  tangents  to  two 
circles  may  be  derived  from  this  theorem  ? 


SIMILAR  FIGURES.  69 


37,  Every  straight  line  cutting  the  sides  of  a  triangle 
(produced  when  necessary)  determines  upon  the  sides  six 
segments,  such  that  the  product  of  three  non-consecutive 
segments  is  equal  to  the  product  of  the  other  three. 

The  line  XYZ  must  cut  either  (i.)  two  sides  of  the  triangle  and 
the  third  side  produced  (Fig.  57),  or  (ii.)  all  three  sides  produced 
(Fig.  58).     The  proof  for  both  cases  is  the  same. 
Draw  CD  II  to  AB.     From  the  similar  triangles, 
.AX^AZ       d  BY_BXm 
CD      CZ'  '        CY     CD 
AXx  BY    AZxBX 

rherefore'         -' 


whence,  AX  X  B  Y  x  CZ  =  AZ  x  BX  x  CY. 

This  theorem  was  discovered  by  Menelaus  of  Alexandria,  about 

80  B.C. 

z 


A  X  B 

Fig.  57.  Fig.  58. 

38.    Prove  the  converse  of  the  last  theorem. ' 

Let  XY  produced  cut  AC  produced  in  a  point  P. 
Then,  AXx  BYx  CP  =  APx  BXx  CY. 

But,  by  the  hypothesis, 

AXx  BYx  CZ  =  AZx  BXx  CY- 
whence,  CP -.  CZ=AP.  AZ- 

whence,       AP-  CP:  AZ  -  CZ=  AP  •  AZ, 
or,  AC :  AC ~  AP :  AZ. 

.-.  AP^  AZ- 
that  is,  P  coincides  with  Z. 


70  GEOMETRY. 


39,  Lines  drawn  through  the  vertices  of  a  triangle,  and 
passing  through  a  common  point,  determine  upon  the  sides 
six  segments,  such  that  the  product  of  three  non-consecutive 
segments  is  equal  to  the  product  of  the  three  others. 

The  common  point  0  may  lie  either  within  the  triangle  (Fig.  59), 
or  without  the  triangle  (Fig.  60).  In  both  cases,  apply  Ex.  36  to  the 
&ACD  and  line  EOF,  and  to  the  A  BCD  and  line  AOE\  then  mul- 
tiply the  results. 

This  theorem  was  first  made  known  by  Ceva  of  Milan,  in  1678. 

c  c 


A  D  B  F  E 

Fig.  59.  Fig.  60. 

40,  Conversely,  if  three  lines  drawn  through  the  vertices 
of  a  triangle  determine  upon  the  sides  six  segments,  such 
that  the  product  of  three  non-consecutive  segments  is  equal 
to  the  product  of  the  three  others,  the  lines  pass  through 
the  same  point. 

The  proof  is  similar  to  that  of  Ex.  38. 

41,  By  me.ans  of  Ex.  40,  prove  Ex..  29,  §  2. 

42,  By  means, of  Ex.  40,  prove  Ex.  32,  §  2. 

43,  By  means  of  Ex.  40,  prove  Ex.  43,  §  2. 

§  14.   NUMERICAL  EXERCISES. 

1.  Find  the  fourth  proportional  to  the  lines  whose  lengths 
are  25  feet,  32  feet,  48  feet. 

2,  Find  the  third  proportional  to  the  lines  whose  lengths 
are  36  feet  and  24  feet. 


SIMILAR    FIGURES.  71 


3.    Find  the  mean  proportional  between  the  lines  whose 
lengths  are  28  feet  and  45  feet. 


4,  In  a  triangle  ABO  the  side  ^45  =  305  feet.    If  a  line 
parallel  to  EC  divides  AC  in.  the  ratio  2  :  3,  what  are  the 
segments  into  which  it  divides  AH? 

Let  x  =  one  segment,  then  305  —  x  =  the  other  ;  and  x  :  305  —  x 
-2:3. 

5,  Upon  two  parallel  lines  six  points  are  taken,  A,  J5,  C 
in  one,  and  A',  B\  C'  in  the  other,  such  that  AB  =  2  inches, 
BC=3  inches,  A'  B'  =  1.24  inches,  JB'C'  =  3.1  inches.    Will 
the  lines  A  A',  BB',  CC'  produced  pass  through  the  same 
point  ? 

Find  if  the  given  numbers  are  in  proportion. 

6,  The  sides  of  a  triangle  are  9,  12,  15  ;  find  the  seg- 
ments of  the  sides  made  by  the  bisectors  of  the  angles. 
(No.  143.) 

7,  If  the  sides  of  a  triangle  are  denoted  by  a,  b,  c,  find 

the  segments  of  c  made  by  the  bisector  of  the  opposite  angle. 
What  do  the  results  become  if  a  —  b  ? 

8,  If  the  acute  angles  of  a  right  triangle  are  30°  and  60°, 
what  is  the  ratio  of  the  segments  into  which  the  bisector  of 
the  angle  of  60°  divides  the  opposite  side  ? 

9,  The  sides  of  a  triangle  are  120,  80,  75.     In  a  similar 
triangle  the  side  homologous  to  120  is  equal  to  90  ;  find 
the  other  two  sides. 

10,  Two  lines  start  from  a  point  A  and  cut  two  parallels. 
The  first  line  cuts  the  parallels  in  B  and  (7;  the  second  line 
in  D  and  E.     If  5(7=  4  feet,  BD=  12  feet,  OE=  18  feet, 
AE=  16  feet,  find  AB,  AD,  DE. 


72  GEOMETRY. 


11.  At  the  ends  of  a  line  AB,  perpendiculars  AC,  BD 
are  erected,  and  in  AB  a  point  0  is  taken,  such  that  the 
angles  AOC,  BOD  are  equal.  If  AB  =  25  inches,  AC= 
13  inches,  BD=  7  inches,  find  OA  and  OB. 

Let  OA  =  a,  O£  =  y.     The  A  ^0(7,  501?  (.%.  61),  are  similar. 


/I  0  B  A  B  G 

Fig.  61.  Fig.  62. 

12,  Given  a  trapezoid  ABCD.     The  legs  DA,  CB  are 
divided,  starting  from  the  points  D,  C,  in  the  ratio  2  :  3, 
and  the  points  of  division  E,  .Fare  joined.    Prove  that  EF 
is  parallel  to  the  bases,  and  compute  the  length  of  EF,  if 
AB  =  3850  feet,  CD  =  1245  feet. 

Let  AD,  BC  produced  meet  in  O  (Fig.  62) ;  then 

O A  :  OB  =  DA  :  CB. 

Also,  DE.  EA=  CF  •.  FB- 

whence,  DA  :  CB  =  EA  -.  FB. 

.-.  OA  •.  OB  =  EA-.  FB. 
.-.EFis  II  to  CD. 
Draw  DFG-  then  BG  -.  DC '=  3  :  2,  and  EF ' :  AG  =  2  :  5. 

13,  If  in  the  trapezoid  ABCD  (Fig.  62)  the  bases  AB, 
CD  are  denoted  by  a,  5,  respectively,  and  the  altitude  by 
h,  find  the  altitude  of  the  triangle  AOB  formed  by  pro- 
ducing the  legs. 

14,  A  tree  casts  a  shadow  60  feet  long  at  the  same  time 
that  a  vertical  rod  3  feet  high  casts  a  shadow  2  feet  long ; 
find  the  height  of  the  tree. 


SIMILAR    FIGURES. 


73 


15,  Show  how  to  find,  by  means  of  similar  triangles,  the 
distance  AX  (Fig.  63).  If  AP  =  200  feet,  OP=  20  feet, 
OQ  -32  feet,  find  A X. 


Fig.  63. 


Fig.  64. 


16,  Show  how  to  find  the  distance  between  two  inac- 
cessible objects  X,  Y.    (Fig.  64.)     If  AX=  4  miles,  A  Y 
=  5  miles,  AB  =  200  feet,  A  C=  250  feet,  J5(7=225  feet, 
find  XY. 

17,  The  perimeters  of  two  similar  polygons  are  280  feet 
and  160  feet.     If  a  side  of  the  first  polygon  is  15  feet,  find 
the  homologous  side  of  the  second. 

18,  Required  the  length  of  a  ladder  which  will  reach  a 
window  24  feet  high,  if  the  lower  end  of  the  ladder  is  10 
feet  from  the  side  of  the  building. 

19,  How  far  apart  are  the  opposite  corners  of  a  floor  12 
feet  by  16  feet? 

20,  If  the  side  of  an  equilateral  triangle  is  a,  find  the 
altitude. 

21,  Find  the  legs  of  a  right  triangle  if  their  projections 
upon  the  hypotenuse  are  2.88  feet  and  5.12  feet. 

x2  :  y2  =  2.88  :  5.12,  and  x*  +  f  =  64. 


74  GEOMETRY. 


22,  The  legs  of  a  right  triangle  are  10  feet  and  24  feet. 
Find  their  projections  upon  the  hypotenuse,  and  the  alti- 
tude upon  the  hypotenuse. 

23,  Given  the  legs  a,  b  of  a  right  triangle,  find  their  pro- 
jections x,  y  on  the  hypotenuse,  and  the  altitude  h  upon 
the  hypotenuse. 

24,  The  hypotenuse  of  a  right  triangle  is  1,  and  the  sum 
of  the  legs  is  1.4  ;  find  the  legs. 

25,  Find  the  three  sides  of  a  right  triangle  if  these  sides 
are  three  consecutive  integral  numbers. 

Denote  the  sides  by  x  —  1,  x,  x  +  1. 

26,  Compute  the  legs  of  a  right  triangle  if  their  ratio  is 
3  :  4,  and  the  hypotenuse  is  40. 

27,  Find  the  sides  of  a  right  triangle  if  their  sum  is  132 
and  the  sum  of  their  squares  is  6050. 

28,  If  in   a  right   triangle   the   hypotenuse   is  25  feet, 
one  leg  is  15  feet ;  find  the  altitude  upon  the  hypotenuse. 

29,  The  legs  of  a  right  triangle  are  3.128  and  4.275; 
compute  to  0.001  of  a  unit  the  segments  of  the  hypotenuse 
made  by  the  bisector  of  the  right  angle. 

30,  The  radius  of  a  circle  is  5  inches.     Find  the  distance 
from  the  centre  to  a  chord  8  inches  long. 

31,  The  radius  of  a  circle  is  r.     What  is  the  length  of  a 
chord  the  distance  of  which  from  the  centre  is  J  r  ?     What 
angle  does  it  subtend  at  the  centre  ? 

32,  If  the  radii  of  two  concentric  circles  are  10  inches 
and  8  inches,  find  the  length  of  a  chord  in  the  larger  circle 
which  touches  the  smaller  circle. 


SIMILAR    FIGURES. 


75 


33,    Two  circles,  the  radii  of  which  are 


intersect  at 


right  angles  ;  find  the  distance  between  their  centres. 

The  tangents  at  the  point  of  intersection  are  perpendicular  to  each 
other,  and  form  with  the  line  of  centres  a  right  triangle. 

34.  The  radii  of  two  circles  are  8  inches  and  3  inches, 
and  the  distance  between  their  centres  is  15  inches.     Find 
the  length  of  their  common  exterior  tangent. 

35.  The  radius  of  a  circle  is  6  inches.     Through  a  point 
10  inches  from  the  centre,  tangents  are  drawn.     Find  the 
lengths  of  these  tangents  and  of  the  chord  of  contact. 

36.  The  distance  between  two  parallel  lines  is  a,  and  the 
distance  between  two  points  A,  B  in  one  of  the  lines  is  26. 
Find  the  radius  of  a  circle  which  passes  through  A  and  B, 
and  touches  the  other  line.     What  is  its  value  if  a  =  b  ? 

37.  The  sides  of  a  triangle  are  8,  9,  13  ;  is  the  greatest 
angle  acute,  right,  or  obtuse  ? 

38.  If  in  an  isosceles  triangle  a  denote  one  of  the  equal 
sides,  and  b  half  the  base,  find  the  radius  of  the  circum- 
scribed circle. 

Draw  the  altitude,  and  drop  a  perpendicular  from  the  centre  of 
the  circle  to  one  of  the  equal  sides  ;  two  similar  triangles  are  formed.- 


39.  In  an  isosceles  trapezoid  let  a  =  the  greater  base, 
b  =  the  other  base,  c=  one  of  the  legs ;  find  the  lengths  of 
the  diagonals. 

The  two  diagonals  are  equal.  (Ex.  51,  §  2.)  Draw  (Fig.  65)  CE 
II  to  DA,  CF  _L  to  AB  ;  in  the  isos.  A  CBE,  FB  =  £  EB  =  J(a  -  6) 
Then  apply  No.  162  to  the  A  ACS, 


7G  GEOMETEY. 

40,  Compute  the  sides  of  a  rectangle,  given  a  diagonal  d 
and  the  perimeter  2p.  When  is  the  problem  possible  ? 
when  impossible? 

Let  x  and  y  denote  the  two  sides;  then  x  +  y  =  p, 
whence, 


If  p2<2d2,  or  p<d\/2,  the  problem  is  possible;  ifp*>2dz,  or 
p  >  cZ\/2,  the  roots  are  imaginary,  and  the  problem  is  impossible.  If 
p  =  dV2,  then  x  =  y  =  $p,  and  the  rectangle  is  a  square. 

41,  Two  chords  AB,  CD  intersect  in  M;   if  A M  =  5 
inches,  BM=  6  inches,   CD  =  11.5  inches,  find  CM  and 
MD. 

x  +y  =  11.5,  xy  =  30. 

42,  Two  chords  AB,   CD  intersect  in  M\    if  A  M  =  4 
inches,  BM=  5  inches,  and  the  difference  between  C'Jf  and 
MD  =  8  inches,  find  CD. 

43,  The  diameter  of  a  circle  is  equal  to  30  feet,  and  is 
divided  into  three  equal  parts  ;  find  the  lengths  of  the  per- 
pendiculars drawn  from  the  points  of  division  and  termi- 
nated by  the  circumference. 

44,  What  must  be  the  distance  of  a  point  from  the  cen- 
tre of  a  circle  (radius  =  r)  in  order  that  a  tangent  drawn 
from  a  point  to  the  circle  may  be  equal  to  three  times  the 
radius  ? 

45,  Through  a  point  P,  exterior  to  a  circle,  a  tangent 
PA  and  a  secant  PBC  are  drawn  ;  if  PB  =  5  inches,  BC 
-  4  inches,  find  PA. 

46,  Find  in  a  line  A B  touching  a  circle  (radius  —  r)  in 
A,  a  point  (7,  such  that  the  exterior  part  CD  of  the  line 
which  joins  C to  the  centre  shall  be  equal  to  i  AC. 


SIMILAR    FIGURES.  77 


Let  x  =  AC  (Fig.  66)  ;  then  -  -  CD,  and  .r2  =  -  1 2r  +  *V  whence, 

^  2\i   27 

a;  =  0  or  f  r. 

47,  The  radius  of  a  circle  equals  2  inches.  Through  a- 
point  ^4,  4  inches  from  the  centre,  a  secant  ABC  is  drawn. 
If  5a=  1  inch,  find  AB. 

Let  re  =  AB  (Fig.  67);  then  x(x+  1)=  12,  whence  a;  =  3  or  -4. 
The  negative  root  must  be  rejected,  or  else  considered  as  the  solution 
of  the  question  obtained  by  changing  x  to  —  x  in  the  equation  x(x+ 1) 
=  12.  The  equation  then  becomes  x2—  #  =  12,  and  belongs  to  the 
question,  find  the  length  of  the  secant  ACif  BC=  1  in. 

C 


Fig.  67. 

48,  In  a  triangle  given  the  sides,  a  =  1551  feet,  b  =  2068 
feet,  c  =  2585  feet  ;  find  the  median  to  the  side  a. 

Ifx  equal  the  required  line,  then  (Ex.  19,  \  13)  z2  +  b2  =  2x*  +  £e2. 
In  this  particular  case  the  labor  of  computing  x  may  be  avoided  by 
observing  that  the  given  numbers  are  equimultiples  of  3,  4,  5,  respec- 
tively ;  therefore  the  triangle  is  a  right  triangle  (No.  161),  and 
B  =  4e.  (Ex.  45,  §2.) 

49,  In  a  triangle  given  the  sides  a,b,c;  find  the  lengths 
of  the  three  medians.     (Ex.  19,  §  13.) 

If  x,  y,  z  denote  the  lengths,  then 


y  = 


By  adding  the  squares  of  these  values,  we  obtain 

0^+2/2  +  22  =  f(a2  +  62  +  c2); 

or,  the  sum  of  the  squares  of  the  medians  is  equal  to  three-fourths  of 
the  sum  of  the  squares  of  the  sides. 


78 


GEOMETRY. 


50.  In  a  .triangle  given  the  sides  a,  b,  c ;  find  the  lengths 
of  the  three  bisectors. 

Let  x,  y,  z  (Fig.  68)  denote  the  lengths  of  the  bisectors  of  the  angles 
opposite  c,  a,  b,  respectively,  and  let  x  divide  the  side  c  into  the  seg- 
ments u,  v.  Then  ab  =  &  +  uv  (Ex.  18,  §  13),  u  :  v  =  a  :  b  (No.  143), 
and  u  +  v  =  c. 

By  eliminating  u  and  v,  we  have 

a  =  ab\(a  +  b}2  -  c2]  =  ab(a  +  b+  c)(a  +  b  -  c) 
(a  +  &)2  (a  +  by 

Let  a  +  b  +  c  =  2p  ;  then  a  +  b  —  c  =  2(p  —  c),  and 
2  Vabp  (p  —  c) 


Similarly, 


a  +  b 


—  a)  . 


b  +  c 


a  +  c 


T 
Fig.  68. 


H 


Fig.  69. 


51.    Given  in  a  triangle  ABC  the  sides  a,  b,  c  ;     find  the 
segments  of  AB  made  by  the  altitude  upon  AB. 

Let  x  =  AH  (Fig.  69) ;  then  c  -  x  = 

Then  Off2  =  a2  -  (c 

also  CZf  =  62  —  re2, 


whence 
and 


2c 


2c 


2c 


SIMILAR    FIGURES.  79 


52,  In  a  triangle  given  the  sides  a,b,c;  find  the  lengths 
of  the  three  altitudes. 

Let  x,  y,  z  denote  the  lengths  of  the  altitudes  upon  c,  a  b,  respec- 
tively. By  the  last  Exercise, 


-  / 
V 


4c2 

2     (26c  +  Z>2  +  c2  -a2)(25c  -  62  -  c2  +a2) 


4c2 
+  c  +  a)(6  +  c  —  a)(a  +  b  — 


4c2 


Let  a  +  6  +  c  =  2  p  ;  then  6  +  c  —  a  =  2  (p  —  a),  (a  —  b  +  c)  =  2  (/>  —  6), 
(a  +  6  —  c)  =  2(p  —  c).    Hence, 

I6p(p-a)(p-t>)(p-c). 

X  —  4c2 


Similarly, 


2Vp(p-a)(p-b)(p-c) 
b 


§15.   Loci. 

1,  Notation.   The  same  notation  is  used  as  in  Chapter  II. 
Also,  PH  denotes  the  perpendicular  from  P  to  L. 

2,  To  find  the  locus  of  points  which  divide  lines  drawn 
from  P  to  L  in  the  ratio  m  :  n. 

The  locus  is  the  line  parallel  to  L  which  divides  PH  in  the  given 
ratio.     Special  case :  PH=^  4  inches,  m  =  3,  n  =  1. 

3,  To  find  the  locus  of  the  ends  of  lines  which  are  drawn 
from  P,  and  are  divided  by  L  in  the  ratio  m  :  n. 

Special  case :  PH=  3  inches,  m  —  3,  n  =  2. 


80  GEOMETRY. 


4,  To  find  the  locus  of  the  ends  of  lines  which  are  drawn 
from  L  to  P,  and  then  produced  so  that  they  are  divided 
by  P  in  the  ratio  m :  n. 

Special  case :  PH=  2  inches,  ra  =  4,  n  =  3. 

^  . 

5,  To  find  the  locus  of  points  the  distances  of  which  from 
two  given  parallels  are  as  in  :  n. 

Special  case :  m  =  2,  n  =  1. 

6,  To  find  the  locus  of  points  the  distances  of  which  from 
two  intersecting  lines  L  and  L!  are  as  m  :  n. 

The  locus  consists  of  two  straight  lines.  Draw  parallels  to  L  and 
Z/,  such  that  their  distances  from  L  and  L'  respectively  shall  be  as 
m  :  n ;  these  parallels  will  intersect  in  points  belonging  to  the  required 
locus.  Special  case :  Z  LL1  =  60°,  m=2,n=l. 

7,  Between  the  sides  of  a  given  angle  a  series  of  parallels 
are  drawn ;  to  find  the  locus  of  points  which  divide  these 
parallels  in  the  ratio  m  :  n. 

Special  case :  m  =  4,  n  =  1. 

A 


Fig.  70. 

8,  Through  P  secants  are  drawn  to  a  circle  K\  to  find 
the  locus  of  points  which  divide  the  entire  secants  in  the 
ratio  m :  n. 

Through  points  of  the  locus  D,  E,  F  (Fig.  70),  draw  lines  parallel 
respectively  to  the  radii  OA,  OB,  00;  these  lines  divide  PO  in  the 
given  ratio,  and  therefore  meet  in  a  point  Q  in  PO.  From  the  similar 
triangles,  QD-.QE-.  QF=  OA-.OB-.OG.  But  OA  =  OB  =  00.  There- 
fore QD=  QE  =  QF.  Therefore  the  locus  is  an  arc  with  Q  as  centre 
and  QD  as  radius. 

Special  cases :  (i.)  Pin  K,  and  m  —  n\  (ii.)  P  within  K,  and  ra  =  2?i. 


SIMILAR    FIGURES. 


81 


9,  To  find  the  locus  of  a  point  such  that  the  sum  of 
the  squares  of  its  distances  from  two  given  points  A,  B  is 
constant. 

The  locus  is  a  circle  having  for  centre  the  middle  point  of  the  lino 
AB.  If  P  (Fig.  71)  denote  any  point  of  the  locus,  k*  the  constant 
quantity,  ra  the  median  PM  of  the  &ABP,  then  AP'*  +  BP*=  k*  ;,  also 
(Ex.  19,  \  13),  ZP2  +  £P2=  2m2  +  2AM*  ; 

j.2  _  2  AM* 


Vj 


,  a  constant  quantity. 


If  k2  <  2  AM2,  or  k  <  AMV2,  there  is  no-locus  ;  if  jfe*  -  2  AM  =  0, 
or  k  =  AMV2,  the  locus  is  reduced  to  a  point. 
Special  case :  AB  =  4  inches,  &2  =  100. 


Fig.  71. 


Fig.  72. 


10.  To  find  the  locus  of  a  point  such  that  the  difference 
of  the  squares  of  its  distances,  from  two  given  points  A,  B 
is  constant. 

The  locus  consists  of  two  straight  lines  perpendicular  to  the  line 
AB.  If  P  (Fig.  72)  is  a  point  of  the  locus,  kz  the  constant  quantity, 
5"  the  foot  of  the  perpendicular  from  P  to  AB,  then  AP*  -  J3P2  =  F  ; 
also  (Ex.  20,  \  13),  AP2  -  BP2  =  2ABx  MH; 

7.2 

,  a  constant  quantity. 


whence, 


MH- 


Therefore,  the  perpendicular  to  AB  erected  at  H  is  one  part  of 
the  locus,  and  the  distance  MH\s  a  third  proportional  to  the  lengths 
2  AB  and  k.  The  other  part  of  the  locus  is  the  corresponding  per- 
pendicular erected  at  the  distance  MH  on  the  other  side  o 

Special  case :   AB  =  6  inches,  P  =  64. 


82 


GEOMETRY. 


11.  Through  P  any  line  PMNis  drawn,  cutting  a  circle 
./Tin  M and  N,  and  P  moves  so  that  the  product  of  the 
segments  PMx  PJV'has  the  constant  value  &2;  to  find  the 
locus  of  P. 


Fig.  73. 


Fig.  74. 


The  locus  is  a  concentric  circle  with  the  radius  OP;  and  the  con- 
stant product  k*  is  called  the  power  of  P  with  respect  to  the  circle  K. 

If  P  is  without  K(Fig.  73),  draw  the  tangent  PT;  then  PMx  PN 
=  PT'2  =  OP2  -  r2.  If  P  is  within  K  (Fig.  74),  draw  APE  _L  to  CD ; 
then  PMx  PN=PAxPB  =  P&  =  r2  -  OP2.  Therefore,  in  both 
cases  OP  is  constant. 

In  case  (i.),  k  =  PT;  in  case  (ii.),  k  =  PA  =  PB.  What  is  the  locus 
if  *  =  r?  if  A  =  0. 


Fig.  75. 

12,  To  find  the  locus  of  a  point  the  distances  of  which 
from  two  given  points  A,  B  are  in  the  ratio  in  :  n. 

Construction.  Through  AB  (Fig.  75)  draw  any  two  parallels  ; 
upon  one  take  AE=  m,  and  upon  the  other  take  BF=  n  and  BO  =  n. 
Draw  EF  cutting  AB  in  C,  and  EG  cutting  AB  produced  in  ,D.  Then 


SIMILAR    FIGURES. 


83 


C,  D  are  two  points  in  the  locus,  and  the  circle  described  upon  CD  as 
diameter  is  the  required  locus. 

Proof.  It  follows  from  the  similar  &ACE,  BOF,  and  ADE,  BDG, 
that  C  and  D  are  points  in  the  locus.  Let  P  be  any  other  point  in 
the  locus.  Join  PA,  PB,  PC,  PD.  By  hypothesis,  PA  :  PB  =  m  •.  n. 
.-.  PA  -.  PB  =  CA  -.  CB  =  DA  :  DB.  .-.  (Exs.  4,  5,  §  13)  PC  bisects 
Z  APS,-  and  PD  bisects  Z  5PJ/!  .-.  CPZ)  =  90°.  .-.  (Ex.  8,  §  5)  the 
locus  of  P  is  the  circle  described  upon  CD  as  diameter. 

Special  cases  :  (i.)  A.B  =  6  inches,  m  =  2n;  (ii.)  .A.Z?  =  4  inches, 
ra  =  n. 


13,  Through  a  fixed  point  J^a  secant  is  drawn  to  a  given 
circle,  and  through  the  intersections  A,  B  with  the  circum- 
ference tangents  are  drawn  intersecting  in  a  point  P.  If 
the  secant  revolves  about  F,  find  the  locus  of  P. 

Let  O  (Fig.  76)  be  the  centre  of  the  circle.  Join  PO,  cutting  AB 
in  (7,  and  draw  PD  _L  to  FO.  Since  P  and  0  are  each  equidistant 
from  A  and  B,  PO  is  _L  to  AB.  The  rt.  A  OPD,  OCF,  having  an 
acute  angle  common,  are  similar.  Therefore, 


But,  in  the  rt.  A  OBP,  OPxOC=OBi  =  r2. 

Therefore,  OD  =  -  ,  a  constant  quantity. 

OF 

Hence,  the  locus  is  a  straight  line  perpendicular  to  FO,  and  pass- 
ing through  the  two  points  of  contact  of  the  tangents  drawn  from  F 
to  the  circle. 


84 


QEOMETBY. 


14,  To  find  the  locus  of  points  such  that  the  tangents 
drawn  from  each  point  to  two  given  circles  shall  be  equal. 

The  locus  is  a  line  perpendicular  to  the  line  of  centres,  and  is  callecj 
the  radical  axis  of  the  two  circles.  Let  P  (Fig.  77)  be  a  point  in  the 
locus.  Then,  PA  =  PB-  whence,  from  the  rt,  A  POA,  PQS, 


or,  "PO2  -'PQ2  =  ~AOt.  -  Q#V  a  constant  quantity. 

This  reduces  the  problem  to  Ex.  12,  and  it  follows  that  the  locus 
is  a  line  perpendicular  to  OQ,  and  cutting  OQ  in  a  point  H,  found  as 
in  Ex.  12.  If  the  circles  are  exterior  to  each  qther^  the  shortest  way 
to  construct  the  locus  is  to  draw  a  common  tangent  TS,  bisect  it  in  C, 
and  then  draw  CH  _L  to  OQ. 


oc.  — 


Fig.  77. 

\  If  the  two  circles  touch  each  other,  the  radical  axis  is  the  common 
interior  tangent;  if  they  cut  each  other,  the  radical  axis  is  the  com- 
mon chord.  For,  if  the  radical  axis  meets  one  of  the  circles,  it  must 
meet  the  other  in  the  same  point,  in  order  that  the  two  tangents 
drawn  from  this  point  may  have  the  same  value,  namely,  zero. 

If  through  P  a  secant  PDE  is  drawn  to  one  of  the  circles,  and  a 
secant  PFG  to  the  other,  then  PDxPE=  PA2,  and  PFxPG  =  P&. 
.-.  PDxPE  =  PFx  PO ;  that  is,  the  radical  axis  is  also  the  locus  of 
points  of  equal  power  with  respect  to  the  two  circles. 

It  is  evident  that  a  circle  with  Pas  centre,  and  PA  as  radius,  will 
cut  both  the  given  circles  at  right  angles  :  so  that  the  radical  axis  is 
also  the  locus  of  the  centres  of  all  circles  which  cut  two  given  circles 
at  right  angles. 

Special  cases  :  (i.)  OQ  =  6  inches,  r  =  3  inches,  r'  =  2  inches;  (ii.) 
OQ  =  6  inches,  r  =  r'  =  2  inches. 


SIMILAR    FIGURES.  85 


15,  To  find   the  locus  of  points  such  that  the  tangent 
drawn  from  each  point  to  a  given  circle  shall  be  equal  to 
the  distance  of  the  point  from  a  given  point  P. 

This  is  a  special  case  of  Ex.  12 ;  for  the  reasoning  is  independent 
of  the  size  of  the  two  circles,  and  holds  true  if  one  of  the  circles  is 
reduced  to  a  point.  Therefore,  the  locus  is  a  line  perpendicular  to 
the  line  which  joins  Pto  the  centre  of  the  given  circle.  If  P  is  with- 
out the  circle,  the  locus  bisects  the  tangent  drawn  fromP;  if  Pis 
within  the  circle,  the  locus  bisects  the  two  tangents  whose  chord  of 
contact  is  bisected  by  P.  (Proof.) 

Special  cases  :  (i.)  OP=  5  inches,  r  =  4  inches ;  (ii.)  OP=  4  inches, 
r  =  4  inches. 

16,  To  find  the  locus  of  points  from  which  two  given 
circles  will  be  seen  under  equal  angles. 

Show  that  the  distances  from  any  point  in  the  locus  to  the  centres 
of  the  two  circles  are  as  the  radii  of  the  circles ;  this  reduces  the 
problem  to  Ex.  12. 

17,  To  find  the  locus  of  the  points  from  which  a  given 
straight  line  is  seen  under  a  given  angle. 

18,  To  find  the  locus  of  the  vertex  of  a  triangle,  having 
given  the  base  and  the  ratio  of  the  other  two  sides. 

19,  To  find  the  locus  of  the  points  in  a  plane  equally 
illuminated  by  two  lights  A,  B  placed  in  the  plane ;  given 
that  the  intensities  of  the  two  lights  at  unit  distance  are  as 
m  :  n,  and  that  the  intensity  varies  inversely  as  the  square 
of  the  distance. 

Special  case  :  AB  =  6  inches,  m  =  2,  n  =  1. 

20,  Through  a  point  A  a  line  is  drawn  meeting  a  given 
circle  in  B  and  C.     In  this  line  a  point  P  is  taken  such 
that  AP  X  AC=  k\     Find  the  locus  of  the  point  P. 

21,  A  given  point  0  is  joined  to  any  point  Min  a  given 
line.     Upon  the  line  OA  a  length  OP  is  taken  such  that 
OP  X  OA  =  Jc\     Find  the  locus  of  the  point  P. 


86  GEOMETRY. 


§  16.   PROBLEMS. 

1,  To  construct  the  mean  proportional  between  two  given 
lines  by  three    different  methods,  and   to  show  that  the 
methods  verify  one  another.     (Nos.  159,  160,  166.) 

2,  Explain  how  a  line  26  inches  long  may  be  divided 
into  three  parts  proportional  to  the  numbers  2,  f  ,  1.     Com- 
pute the  three  parts. 

To  divide  a  line  AB  in  a  point  (7  so  that, 

3,  ACxCB  =  k\  4,    ABxAC=lc\ 

5,  To  construct  two  lines,  given  their  sum  and  their  ratio. 

6,  To   construct  two  lines,   given  their  difference   and 
their  ratio. 

7,  To  cut  from  two  lines  equal  lengths  such  that  the 
remaining  parts  shall  have  the  ratio  m  :  n. 

8,  To    produce    a    line   AB  to   a   point    0  such    that 


9,  To  construct  a  line,  given  the  greater  segment  of  the 
line  divided  in  extreme  and  mean  ratio. 

10,  To  divide  a  line  AB  harmonically  in  a  given  ratio 
m\n\  that  is,  to  find  a  point  C  in  AB  and  a  point-  D  in 
'AB  produced  such  that  AC  :  BC=  AD  :  BD  ==  m  :  n. 

See  solution  of  Ex.  12,  §  15,  or  apply  Nos.  143  and  144.  A  and 
B  are  called  conjugate  points  ;  likewise,  C  and  D.  Examine  the  case 
when  m  =  n. 

11,  Given  in  the  harmonic  division  of  a  line  three  points 
A,  B,  C;  to  find  the  fourth  point  D. 

12,  To  construct  a  triangle  similar  to  a  given  triangle  by 
applying  Ex.  29,  §  13. 

Special  case  :  ratio  of  similitude  =  £. 


SIMILAR    FIGURES.  87 


13,  To   draw  the  common  tangents  to  two   circles  by 
applying  the  properties  of  their  centres  of  similitude. 

14,  To  draw  a  line  from  P  to  L  which  shall  be  to  the 
perpendicular  dropped  from  P  to  L  as  m  :  n. 

Given  an  angle  BA  O  and  a  point  P;  through  P  to  draw 
a  line  meeting  the  sides  of  the  angle  in  X  and  Y,  so  that, 

15,  AX:AY=m:n.  16,    PX:PY-=m\n. 

In  Ex.  15  and  16  take  P(i.)  within  the  angle,  and  also  (ii.)  with- 
out the  angle. 

17,  Three  given  lines  meet  in  a  point  P;  through  Pto 
draw  a  line  such  that  the  two  segments  made  by  the  given 
lines  may  have  a  given  ratio. 

Through  any  point  in  the  middle  one  of  the  given  lines,  draw  a 
line  so  that  its  segments  may  have  the  given  ratio ;  then,  through  P 
draw  a  parallel. 

Given  P  and  L ;  to  find  a  point  X  such  that  PX  shall 
be  cut  by  L  in  a  given  ratio,  and 

18,  X  shall  be  at  the  distance  a  from  P,. 

19,  X  shall  be  at  the  distance  a  from  L\. 

20,  X  shall  be  equidistant  from  Pl  and  P2. 

21,  X  shall  be  equidistant  from  LI  and  L2. 

22,  In  one  side  of  a  given  triangle  to  find  a  point,  the 
distances  of  which  from  the  other  sides  shall  have  a  given 
ratio. 

23,  Within  a  given  triangle  to  find  a  point  the  distances' 
of  which  from  the  three  sides  shall  be  as  the   numbers 
m  :  n  :  p. 

24,  Between  two  parallel  tangents  to  a  circle,  to  draw  a 
third  tangent  so  that  it  shall  be  divided  by  the  point  of 
contact  in  a  given  ratio. 


88  GEOMETRY. 


25,  Given  L,  I/lt  K\  to  construct  a  circle  having  a  given 
radius,  touching  K,  and  having  the  distances  from  its  centre 
to  L  and  LI  in  a  given  ratio. 

26,  Given  P,   L  II  to  I/lt   and   L2  cutting   L  and   Lv ; 
through  P  to  draw  a  line  from  which  the  given  lines  shall 
cut  two  segments  having  a  given  ratio. 

27,  In  K  given  P  and  Pl\  to  place  in  K  a  chord  of 
given  length  so  that  the  distances  from  P  and  Pl  to  the 
chord  shall  be  as  3  :  1. 

To  find  a  point  X  such  that  the  distances  from  two  given 
points  shall  have  a  given  ratio,  and 

28,  X  shall  be  at  the  distance  a  from  L. 

29,  X  shall  be  equidistant  'from  L  and  Lv 

30,  The  tangent  from  X  to  K  shall  have  a  given  length. 

31,  The  distances  from  X  to  two   other   given   points 
shall  also  have  a  given  ratio. 

32,  In  a  given  triangle,  to  find  a  point  the  distances  of 
which  from  the  three  vertices  shall  be  as  the   numbers 

m  :  n :  p. 

33,  To  find  a  point  from  which  the  lengths  AB,  EC, 
CD  taken  in  a  straight  line  shall  be   seen  under    equal 
angles.     (No.  143,  and  §  15,  Ex.  12.) 

34,  To  find  a  point  from  which  three  given  circles  will 
be  seen  under  equal  angles. 

The  distances  of  the  required  point  from  the  centres  of  the  circles 
are  proportional  to  the  respective  radii  of  the  circles.     (Ex.  16,  g  15.) 

35,  In  a  diameter  of  a  circular  billiard  table  are  placed 
two  balls ;  to  find  by  construction  in  what  direction  one  of 
the  balls  must  be  struck  centrally  in  order  that  it  may  hit 
the  other  after  first  striking  the  side  of  the  table. 


SIMILAR    FIGURES. 


89 


36,  To  find  the  point  which  is  equally  illuminated  by 
three  lights  situated  in  the  same  plane.     (Ex.  10,  §  15.) 

37,  To  inscribe  a  square  in  a  semicircle. 

Suppose  the  problem  solved,  and  ABCD  (Fig.  78)  the  square  re- 


quired.    First  prove  that  OA  —  OB  =  • 

•  then  draw  EH  JL  to  EF, 

and  meeting  OD  produced  in  H  ;  we  have  =  Q=  =  -,  whence 
EH=20E. 

H  \ 

\ 
\ 

V  —  Y 

C                            M 

e/\  --"""  "i 

/\  /^\        / 

>"'\           1 

(     V     \        />" 

'      A    ! 

E           A        0        B         F                  A 
Fig.  78. 

D            H    E      B               N 
Fig.  79. 

38,  To  inscribe  a  square  in  a  given  triangle. 

Suppose  the  problem  solved,  and  DEFG  (Fig.  79)  the  inscribed 
square.  Draw  CM  II  to  AS,  ami  let  AF  produced  meet  CM  in  M\ 
then  GF:  CM=AF:  AM.  Draw  MN  ±  to  AB,  and  OH  A.  to  AB  ; 
then  FE :  MN=  AF  -.  AM.  Now  GF  =  FE •  .-.  CM '=  MN  =  CH. 
Therefore,  construct  a  square  upon  CH  as  one  side,  and  the  line  AM 
will  determine  the  point  F, 

39,  To  inscribe  in  a  triangle  a  rectangle  similar  to  a 
given  rectangle. 

The  solution  is  like  that  of  Ex.  38  ;  upon  CH  (Fig.  79)  construct  a 
rectangle  similar"  to  the  given  rectangle. 

40,  To  inscribe  in  a  semicircle  a  rectangle  similar  to  a 
given  rectangle. 

41,  To  inscribe  in  a  circle  a  triangle  similar  to  a  given 
triangle. 

42,  To  circumscribe  about  a  circle  a  triangle  similar  to 
a  given  triangle. 


90  GEOMETRY. 


In  the  triangle  ABC  to  draw  a  parallel  to  AB,  meeting 
A  O  in  Xand  EG  in  Y,  so  that, 

43,    A£:XY=m:n.      44,    AC:  XY=  XY  \  CX. 

From  the  similar  A  ABC,  XYC  we  have  AB  :  AC  =  XY  :  CX; 
.-.  AB-.  AC  =  AC:  XY. 

45,  AB:  XY=  XY:  CX. 

46,  BC  :  XY=  XY:  CX. 

47,  AB:XY=AX:BY. 

48,  AX:  XY=  XY:  BY. 

Through  C  draw  a  line  II  to  AS;  produce  AY&nd  BXto  meet  it. 

49,  Given  P  within  K;  through  P  to  draw  a  chord  XY 
so  that  PX:  P  Y~-=  m  :  n.     If  r  =  4  inches,  PO  =  2  inches, 
between  what  limits  must  the  ratio  m  :  n  lie  in  order  that 
the  problem  may  be  possible  ? 

Draw  A  Y  II  to  OX  and  meeting  OP  in  A  ;  then  OP:  OA  =  m  :  n, 
also  r  :  -AF=  m  :  n. 

50,  Given  P  in  the  arc  AB  of  j&T;  to  draw  a  chord  PX 
which  shall  be  divided  by  the  chord  AB  in  the  ratio  m  :  n. 

Draw  XC  II  to  AB  and  meeting  PO. 


51.    Given  K,   a  chord  AB,  P  in  ^4.5  ;  through  P  to 
draw  a  chord  XYso  that  J.JT:  BY=m  :n. 


52,  Given  JTand  a  chord  ^4.J5  ;  to  find  a  point  Xin  K 
so  that  -4  JT  :  BX  =  m  :  n. 

53,  In  K  to  draw  a  chord  which  shall  be  divided  by 
two  given  radii  into  three  equal  parts. 

54.  Given  P  in  K\  from  P  to  draw  two  chords  PX, 
PFso  that  PX:  PY=  m  :  n,  and  XY  shall  be  a  diameter. 

Through  P  draw  a  line  JL  to  PO  ;  two  similar  A  are  formed. 

55.  In  a  given  arc  AB  to  find  a   point  X  such  that 

BX=V. 


SIMILAR    FIGURES.  91 


Given  P  without  K\  through  P  to  draw  a  secant  PXYt 
so  that, 

56,  PX:  XY  =  XY:  PY. 
Draw  a  tangent  from  P  to  the  circle. 

57,  XY=2PX.  59,    PXxXY=k\ 

58,  PXxXY=tf.          60,    PY:  PX=  PX:  XY. 

Draw  a  tangent  PA,  also  BX  II  to  AY;  then  PA' is  divided  by  B 
in  extreme  and  mean  ratio  ;  also  from  the  similar  A  PAX,  PBX  we 
have  PX'1-  =  PAx  PB. 

61,  Given  ./Tand  two  radii  OA,  OB;  to  draw  a  tangent 
CXD,  bisecting  Kin  Xand  meeting  OA,  OB  produced  in 
C,  D,  so  that  CX  :  XD  =  m  :  n. 

Draw  AE  II  to  CD,  meeting  OJf  in  E  and  OD  in  P",  and  draw 
GG  II  to  OD  meeting  OJ.  in  (7 ;  then  AO  :  GO  =  m  :  n. 

62,  Given  jSTand  -ffi  intersecting  in  P;  through  P  to 
draw  a  line  so  that  the  chords  intercepted  by  the  circles 
shall  be  as  m  :  n. 

A  line  through  P  perpendicular  to  the  required  line  will  divide 
the  line  of  centres  in  the  ratio  m  :  n. 

63,  In  L  to  find  a  point  X  from  which  the  tangents 
drawn  to  J^and  K^  shall  be  equal. 

64,  To  find  a  point  X  from  which  the  tangents  drawn 
to  three  given  circles  shall  be  equal. 

65,  To  construct  a  circle  having  a  given  radius,  and  cut- 
ting two  given  circles  at  right  angles. 

66,  To  construct  a  circle  which  shall  pass  through  P  and 
cut  _STat  right  angles  in  Pr. 

67,  To  construct  a  circle  which  shall  pass  through  P  and 
cut  JTand  Kr  at  right  angles. 

68,  To  construct  a  circle  which  shall  cut  three  given 
circles  at  right  angles. 


02  GEOMETRY. 


§  17.   THE  METHOD  OF  SIMILAR  FIGURES. 

1,  This  method  of  constructing  figures  is  based  upon  the 
fact  that  the  shape  of  a  figure  is  determined  by  certain 
angles  or  the  ratios  of  certain  lines.     First  construct  a  fig- 
ure similar  to  the  figure  required,  and  then  construct  the 
required  figure  by  means  of  the  general  theorem  that  in 
similar  figures  homologous  lines  are  proportional. 

2,  The  shape  of  a  triangle  is  determined  by : 
(i.)  two  angles ; 

(ii.)  the  ratio  of  two  sides  and  the  included  angle  ; 
(iii.)  the  ratios  of  the  three  sides ; 
(iv.)  the  ratios  of  the  three  altitudes  ; 
(v.)  the  ratio  of  an  altitude  to  the  corresponding  base, 
and  the  angle  at  the  vertex. 

3,  This  method  is  also  applicable  to  cases  in  which  the 
shape,  not  of  the  required  figure,  but  of  an  auxiliary  trian- 
gle, is  determined  by  the  given  parts ;  as,  for  example,  by : 

(i.)  one  angle  ; 

(ii.)  the  ratio  of  an  altitude  to  one  of  the  adjacent  sides  ; 
(iii.)  the  ratio  of  an  altitude  to  the  corresponding  median  ; 
(iv.)  the  ratio  of  the  radius  of  the  circumscribed  circle 
to  one  side. 

4,  The  notation  used  in  the  following  exercises  is  the 
same  as  in  Chapter  II. 

5,  To  construct  a  triangle,  given  a,  ft,  h-\-  m. 

Analysis.  Any  A  DEO  (Fig.  80),  in  which  Z  CDE  =  a,  Z  OED  =  0, 
will  be  similar  to  the  triangle  required.  If  in  this  triangle  we  draw 
the  altitude  GF  and  the  median  CO,  then  will  each  side  of  the  re- 
quired triangle  be  a  fourth  proportional  to  three  known  lengths, 
£/F  +  CG,h  +  m,  and  the  homologous  side  of  the  A  DEC.  The  prob- 
lem is  then  reduced  to  the  construction  of  a  triangle  from  twp  angles 
and  one  side. 


SIMILAR    FIGURES. 


93 


Construction.  Upon  any  line  DE  construct  a  A  DEC,  making 
Z  CD  E  =  a,  Z  CED  =  0.  Draw  OF  ±  to  DE,  bisect  DE  in  Q,  and 
join  CG.  Produce  OF  to-  L,  making  FL  =  CG,  and  upon  CL  take 
CK  =  h  +m.  Join  LD,  and  through  K  draw  JiTA  II  to  LD,  cutting 
CD  (or  C!Z)  produced)  in  A.  Then  draw  AB  II  to  D.fi',  and  meeting 
CE(or  CE  produced)  in  B.  A  ABC  is  ,the  triangle  required. 


Jf  is  middle 


Proof.     Z  CA5  =  Z  <7D#  =  a, 
Let  AB  cut  CF  in  H,  CG  in  Jf. 
point  of  AB. 


=  Z 

Gff  is  JL  to 


But 


Whence, 


CH:  CF=  CM-  CG^CA:  CD, 
CH+  CM:  CF+  CG=CA:  CD, 
CA-.CD=CK:  CL. 
CH+  CM:  CF+CG=CK:  CL. 
CF+  CG  -  CL,  and  CK=  h  +  m. 

M:CL  =  h  +  m:  CL. 
CH  +  CM  =  h  +  m. 


Discussion.   There  is  always  one,  and  only  one,  solution. 

C 


h  +  m 


X 


Fig.  80. 

6,    To  construct  a  triangle,  given  h :  a,  c  +  h,  a. 

Analysis.  Let  ABC  (Fig.  80)  be  the  triangle  required,  CH  its  alti- 
tude h.  The  given  ratio  h  :  a  determines  the  shape  of  the  rt.  A  CBH. 
Therefore,  if  we  construct  two  lines  m  and  n,  so  that  m:  n  =  h  -.a,  and 
then  construct  a  rt.  A  CEF,  making  C!F=  m,  CE=  n,  the  rt.  A  CEF 
will  be  similar  to  the  rt.  A  CBH;  and  if  we  draw  through  C  a  line 
CD,  making  Z.CDE  =  o,  we  shall  have  a  A  CDE  similar  to  the  tri- 
angle required.  The  remainder  of  the  solution  is  precisely  similar  to 
that  of  the  last  exercise. 


94 


GEOMETRY. 


7,    To  construct  a  triangle,  given  the  three  altitudes  h, 

,  A,. 

Analysis.    By  Ex.  2,  \  13,  a  :  c  =  h  •.  ha,  and  6  :  c  =  h  :  hi,  ;  hence, 


that  is,  the  sides  of  the  triangle  required  are  proportional  to  the 
numbers 


and  any  A  DEO  (Fig.  81),  whose  sides  are  proportional  to  these 
numbers,  will  be  similar  to  the  triangle  required.  By  drawing  CF 
_L  to  DE,  taking  GH=  h,  and  drawing  through  H,  AB  II  to  DE,  we 
obtain  the  required  A  ABC. 


Fig.  81. 

To  construct  a  triangle,  given  : 

8,  Rt.  A,  a,  m.  16,   a,  ft,  c  —  h. 

9,  Rt.  A,  a,  c-\-h. 

10,  Rt.  A,  a,c  —  h. 

11,  Rt.  A,  a,  p. 

12,  Isos.  A,  y,  c  +  h. 

13,  Tsos.  A,  y,  <?-fr. 

14,  Isos.  A,  a,  a  — p. 

15,  Isos.  A,  a,  r-\-p. 


17.  a,  ftr-fp. 

18.  a,  ft,  T  —  p. 

19.  a,  p,  r  +  A. 

20.  a,/?,  £  +  *. 

21.  a  :  5,  c  -f  A,  y. 

22.  a  :  b,  c  —  h,  y. 

23.  a:  b,  r-}-  p,  y. 


SIMILAR    FIGURES.  95 


24.  a  :  b,  r  -f-  h,  y.  50,  h  :  771,  a  -f-  b,  a. 

25.  a:b,h-i~p,y.  51,  A  :  m,  a  -f  e,  a. 

26.  a:b,  c+t,y.  52.  A  :  m,  h  -f  c,  a. 

27.  a  :  6,  c  -f-  m>  7-  53,  A  :  m,  m  -f  <?,  a. 

28.  a  :  b,  m  -f-  A,  y.  54.  h  :  m,  a,  ft. 

29.  a  :  b,  b  :  c,  h.  55.  h:m,  a,  c. 

30.  a:b,b:  c,  p.  56.  h  :  m,  c,  ft. 

31.  Aa  :  A5,  r,  y.  57.  h  :  a,  a:  c,  ha. 

32.  Att  :  A6,  p,  y.  58,  h:a,  a:  c,  t. 

33.  Aa  :  A6,  A,  y.  59,  A  :  a,  a  :  c,  r  -f  p. 

34.  Aa  :  hb,  m,  y.  60,  r  :  a,  c,  ft. 

35.  h  :  Aa,  h  :  hby  r.  61.  r  :  a,  p,  ft. 

36.  h:ha,h:  hb,  p.  62.  r  :  a,  b  +  c,  /?. 

37.  h  :  ha,  h  :  hb,  m.  63.  r  :  a,  b  :  c,  ha. 

38.  A  :  Aa,  h  :  hb,  t.  64,  r:a,b:c,r  +  p. 

39.  A  :  a,  h  :  6,  c.  65.  t:h,c,  y. 

40.  h:  a,  h:  b,  m.  66,  t:  h,  r,  a. 

41.  h:a,  h:b,  L  67.  * :  A,  /o,  £. 

42.  A  :  a,  5,  a.  68.  t\h,p  —  q,  a. 

43.  A  :  a,  c,  a.  69.  t :  A,  a  —  b,  y. 

44.  A:  a,  b  +  c,  a.  70.  a:b,a  —  ft,h. 

45.  A  :  a,  a  +  c,  6.  71.  a:b,a  —  ft,  r. 

46.  A  :  a,  c,  r.  72.  a:b,  a  —  ft,  p. 

47.  A  :  c,  a  -f  5,  y.  73.  a  :  (p  T  gr),  c,  0. 

48.  A  :  c,  a  —  6,  y.  74.  a:(p-q),b  +  c,  ft. 

49.  A  :  c,  A  +  p,  y.  75.  a  +  6  :  p  —  £,  c,  0. 


96  GEOMETRY. 


76,  a  +  b  :p  —  q,  h,  ft.  83,  p  :  q,  a,  r. 

77,  p  :  q,  h,  a.  84,  h  :  in,  u  —  v,  a. 

78,  p  :  q,  h,  y.  85,  h  :  m,  a      ft,  t. 

79,  p  :  q,  p,  y.  86,  h  :  m,  a  -f  b,  y. 

80,  p  :q,b  +  c,  a.  87,  c:h,a-ft,a  +  b. 

81,  p  :  q,  a  -  ft,  h.  88,  c  :  h,  a  -  ft,  a  -  b. 

82,  p  i  q,  h,  t. 

To  construct  a  rectangle,  given  : 

89,  a:f,b+f.  92,  a:b,f. 

90,  a  if,  a  +  b.  93,  aibj+a. 

91,  a  if,  a  +  b  -f.  94,  aib,f—  a. 

To  construct  a  parallellogram,  given  : 

95,  a  i  b,  a,  h.  98,  a  i  b,  a,f+g. 

96,  a:  b,  a,  a  -f-  A.  99,  a  :  b,  a,  a  +/• 

97,  a-b,a,a-h.  100,  a  :  6,  a,  A  +/. 

To  construct  a  chord  quadrilateral,  given  a  i  b,  and  also, 

101,  r,  c,  a.  104,  r,f,  a.  107,  r,  a,  ft. 

102,  C,  C?,  a.  105,    c,f,  a.  108,    C,  a,  0. 

103,  C,  #,  a.  106,  /  </,  a.  109,  /,  a,  0. 

To  construct  a  chord  quadrilateral,  given  : 

110,  a,  c,  a,  6.          Ill,  a,  c,  rf,  0.         112,  a,  c,/,  0. 

To  construct  a  tangent  quadrilateral,  given  : 

113,  /,  a,  £,  y.  115,    a  i  b,  p,  a,  y. 

114,  a,  b,  c  i  d,  a.  116,  a  i  b,  c,  a,  y. 

To  construct  a  quadrilateral,  given  the  four  angles  which 
the  diagonal /forms  with  the  sides,  and  also 
117,  a  +  b.        118,  (j.        119,  f+g.        120, 


SIMILAR    FIGURES. 


97 


§  18.     THE  PROBLEM  OF  APOLLONIUS. 

To  construct  a  circle  which  shall  touch  three  given  circles. 

This  problem  was  first  solved  by  Apollonius,  a  famous  Greek 
geometer,  who  lived  about  200  B.C.  It  is  the  last  of  the  following 
series  of  ten  closely  related  problems. 

In  these  problems  P,  Plt  P2  denote  given  points ;  L,  Llt  L2  given 
lines  ;  K,  K^  K2  given  circles  ;  0,  Ov  02  their  centres  ;  r,  rlt  r2  their 
radii ;  X  the  centre  of  the  required  circle. 

1,    To  construct  a  circle  through  P,  Plt  P2. 

Solution  in  all  text-books.     When  is  the  problem  impossible  ? 


A  L  Y 

Fig.   82. 

2,  To  construct  a  circle  through  P,  Plt  and  touching  L. 
Analysis.    Let  the  line  PP,  (Fig.  82)  cut  L  in  A,  and  the  required 

circle  -rot  L  in  Y.  Then  AY'2  =  AP  X  -AP,,  whence  F  may  be 
found.  One  locus  of  X  is  the  line  through  Y  and  ±  to  A  Y,  and  a 
second  locus  is  the  bisector  of  the  angle  PA  Y. 

Discussion.  Since  A  Y  may  be  taken  from  A  upon  L  in  two 
directions,  there  are  in  general  two  solutions. 

Special  case :  PP1  II  to  L. 

3,  To  construct  a  circle  through  P,  P1;  and  touching  K. 
Analysis.    Let  the  required  circle  touch  K  in  Y  (Fig.  83),  and  the 

common  tangent  through  Y  meet  PPj  in  M,  and  MAB  be  any  secant 
through  M,  cutting  K  in  A  and  B  \  then 

Jl/F2  =  J/Px  MPl  =  MAx  MB.  (No.  166.) 

Hence,  a  circle  may  be  described  through  P,  Pls  vl,  B  (Ex.  10,  g  13). 


98 


GEOMETRY. 


Conversely,  to  determine  F,  describe  any  circle  through  Pand  Plf 
let  it  cut  K  in  A  and  £,  produce  AB  to  meet  PPl  in  M,  and  through 
M  draw  a  tangent  to  K.  When  Fis  known,  the  problem  is  reduced 
to  Ex.  1.  Fig.  83  shows  the  construction. 

Discussion.  The  two  tangents  through  M  to  K  furnish,  in  general, 
two  solutions.  Special  cases:  (i.)  P without  K,  and  P1  in  the  tan- 
gent from  P  to  K\  (ii.)  P  in  the  circumference  of  K  and  P1?  (a)  in, 
(6)  within,  (c)  without,  the  circumference  of  K;  (iii.)  P  and  Pl  within 
K;  (iv.)  P,  P^  and  0  in  a  straight  line  ;  (v.)  P  and  P!  equidistant 
from  K. 


4,    To  construct  a  circle  through  P,  and  touching  L,  L^ 

Analysis.  Bisect  that  angle  formed  by  L  and  Lr  (Fig.  84),  in  which 
P  lies ;  draw  a  perpendicular  PS  from  P  to  the  bisector ;  produce 
PB  to  Q,  making  BQ  =  BP\  then  Q  is  a  second  point  in  the  required 
circumference,  and  the  problem  is  reduced  to  Ex.  2. 

Second  Analysis.  About  any  point  0  (Fig.  84),  in  the  bisector 
as  centre,  describe  a  circle  touching  L  and  L^ ;  the  point  A  is  a  centre 
of  similitude  of  this  circle  and  the  required  circle.  (Ex.  36,  §  13.)  Let 
^IPcut  this  auxiliary  circle  in  D  and  E\  then  OD  or  OE  is  parallel 
to  the  radius  of  the  required  circle  drawn  to  P.  (Ex.  33,  $  13.) 

Discussion.  In  general,  there  are  two  solutions.  Special  cases  : 
(i.)  L  II  to  Zj ;  (ii.)  Plies  in  L-  (iii.)  Plies  in  the  bisector  AO. 


SIMILAR    FIGURES. 


99 


5,    To  construct  a  circle  through  P,  and  touching  L,  K. 

Analysis.  Let  the  required  circle  touch  L  in  Y  (Fig.  85),  and  K 
in  Z.  Produce  YZ  to  meet  K  in  A,  join  AO,  and  produce  it  to  meet 
K  a  second  time  in  B  and  L  in  C.  Join  BZ  and  AP,  and  let  AP 
meet  the  required  circle  a  second  time  in  Q. 

A 


.  .:  AOis  II  to  XY.  But  XY 
is  ±  to  L  ;  /.  40  is  -L  to  L.  Also,  ££  is  _L  to  A  Y-  .-.  the  A  ACY} 
ABZ  are  similar  ;  whence,  AB  x  4(7  =  AZx  A  Y.  Also,  4Q  X  AP 
=  AZxAY($o.  165).  Hence,  AS  X  A0=  AQ  X  AP.  /.the  cir- 
cumference which  passes  through  B^_C,  and  Pmust  also  pass  through 
Q.  (Ex.  10,  §  13.)  By  describing  this  circumference,  Q  is  found,  and 
the  problem  is  then  reduced  to  Ex.  2. 


100 


GEOMETRY. 


Discussion.  If  L  does  not  cut  K,  and  P  lies  without  K,  as  shown 
in  Fig.  85,  there  are  four  solutions,  two  of  which  are  obtained  by 
joining  P  to  A,  and  the  other  two  by  joining  P  to  B.  There  are 
numerous  special  cases  for  different  relative  positions  of  P,  L,  and  K. 


6,  To  construct  a  circle  through  P,  and  touching  K,  K\. 
Analysis.   Let  the  required  circle  touch  K,  K^  in  Y,  Z  (Fig.  86). 

Produce  YZ  to  meet  00l  in  M.  Z  OYB  =  /.  XYZ  =  Z  XZY  = 
/.  AZOV  =  Z  O^Z,  /.  OF  is  II  to  0VA,  .'.  M  is  the  direct  centre  of 
similitude  of  K  and  JT,  (Ex.  32,  §  13),  /.  MD  X  MC  =  MYx  MZ 
(Ex.  35,  $  13).  Join  PM,  cutting  the  required  circle  a  second  time 
in  Q-  then  MPxMQ=  MYx  MZ(No.l65),  .'.  MPx MQ=MDxMC, 
whence  it  follows  that  Q  lies  in  the  circumference  passing  through 
the  three  known  points  P,  D,  0.  By  describing  this  circumference 
and  joining  Pto  M,  the  direct  centre  of  similitude  of  K  and  Kv,  the 
point  Q  is  found,  and  the  problem  will  then  be  reduced  to  Ex.  3. 

Discussion.  The  maximum  number  of  solutions  is  four.  In  dis- 
cussing the  numerous  special  cases,  the  position  of  P  relatively  to  the 
common  tangents  of  .ZTand  K^  must  be  considered. 

7,  To  construct  a  circle  touching  L,  Zl5  Z2. 

See  Ex.  66,  \  7.  According  to  the  relative  positions  of  the  lines 
the  number  of  solutions  is  four,  two,  or  none. 

8,  To  construct  a  circle  touching  L,  L^  K. 

Analysis.  Suppose  the  problem  solved,  and  a  concentric  circle 
described  with  X  as  centre  and  XO  as  radius  (Fig.  87) ;  this  cii'cle 
must  touch  parallels  to  L  and  Zls  respectively,  drawn  at  the  distance 
r  from  L  and  Lr  Therefore  X  is  found  by  constructing  this  con- 
centric circle,  as  explained  in  Ex.  4. 


SIMILAR   FIGURES.  101 


Discussion.  The  maximum  number  of  solutions  is  8.  In  discuss- 
ing special  cases,  the  positions  of  L  and  L±  relatively  to  -BTmust  be 
considered. 


-;>'       '   '  •  *•** 

Fig.  87. 

9.  To  construct  a  circle  touching  L,  K,  K^. 

Analysis.  Suppose  the  required  circle  constructed,  and  also  a 
concentric  circle  with  X  as  centre  and  XO  as  radius ;  this  circle  must 
touch  :  (i.)  a  parallel  to  L,  drawn  at  the  distance  r  from  L  •  (ii.)  a 
circle  with  Ol  as  centre  and  rl  —  r  as  radius.  Hence,  JTis  found  "by 
constructing  this  concentric  circle,  as  explained  in  Ex.  5. 

Discussion.  The  use  of  rt  —  r  as  radius  of  the  auxiliary  circle  may 
give  four  solutions,  and  the  use  of  rx  +  r  four  more  solutions. 

10,  To  construct  a  circle  touching  K,  K±,  K2. 

Analysis.  A  concentric  circle,  with  XO  as  radius,  must  touch  the 
circle  with  Ol  as  centre  and  rl  —  r  as  radius,  and  also  touch  the  circle 
wifh.  02  as  centre  and  r2  —  r  as  radius.  By  constructing  this  concen- 
tric circle,  therefore,  as  explained  in  Ex.  0,  X  will  be  determined. 

Discussion.  By  using  for  the  radii  of  the  two '  auxiliary  circles 
about  Ol  and  02  as  centres  different  combinations  of  the  four  values, 
ri  ~  r>  ri  +  r>  r2  ~  r>  r2  +  r>  the  maximum  number  of  solutions  is 
eight. 


102  GEOMETRY. 


CHAPTER    IV. 

EQUIVALENT    FIGURES. 


§  19.    THEOREMS. 

L    All  parai-ielograms  having  equal  bases,  and  contained 
between  two  parallel  lines,  are  equivalent. 

£•    Tv.ro  triangles  having  two  sides  equal,  each  to  each, 
and  the  included  angles  supplementary,  are  equivalent. 

3,  Two  parallelograms  having  two  adjacent  sides  equal, 
each  to  each,  and  the  included  angles  supplementary,  are 
equivalent. 

4,  Every  straight  line  drawn  through  the  centre  of  a 
parallelogram  divides  it  into  two  equivalent  parts.     What 
kind  of  figures  are  the  two  parts  ? 

5,  Every  straight  line  drawn  through  the  middle  point 
of  the  median  of  a  trapezoid,  and  cutting  the  two  bases, 
divides  the  trapezoid  into  two  equivalent  parts.     Is  the 
theorem  also  true  if  the  line  cuts  the  legs  instead  of  the 
bases  ? 

6,  In  every  trapezoid  the  triangle  which  has  for  base  one 
of  the  legs  of  the  trapezoid,  and  for  vertex    the    middle 
point  of  the  opposite  side,  is  equivalent  to  one-half  of  the 
trapezoid. 

7,  The  sum  of  the  areas  of  two  opposite  triangles,  formed 
by  joining  a  point  within  a  parallelogram  to  the  four  ver- 
tices, is  equal  to  one-half  of  the  area  of  the  parallelogram. 


EQUIVALENT    FIGURES.  103 

8,  The  area  of  a  trapezoid  is  equal  to  the  product  of  one 
of  its  legs,  and  the  distance  of  this  leg  from  the  middle 
point  of  the  opposite  side. 

9,  The  triangle  whose  vertices  are  the  middle  points  of 
the  sides  of  a  given  triangle  is  equivalent  to  one-fourth  of 
the  given  triangle. 

10,  The  figure  whose  vertices  are  the  middle  points  of 
the  sides  of  any  quadrilateral  is  a  parallelogram,  and  is 
equivalent  to  one-half  of  the  quadrilateral. 

Hi    Mutually  equiangular  parallelograms  have  the  same 
ratio  as  the  products  of  two  adjacent  sides. 

12,  Similar  polygons  are  to  each  other  as  the  squares  of 
two  homologous  diagonals. 

13,  If  AJ3Cis  a  right  triangle,  C  the  vertex  of  the  right 
angle,  BD  a  line  cutting  AC  in  D,  then  J3D2-\-AC2  = 
A&  -f  DC\ 

14,  If  upon  the  sides  of  a  right  triangle,  as  homologous 
sides,  any  three  similar  figures  are  constructed,  the  figure 
constructed  upon  the  hypotenuse  is  equivalent  to  the  sum 
of  the  figures  constructed  upon  the  legs. 

15,  The  square  constructed  upon  the  sum  of  two  lines  is 
equivalent  to  the  sum  of  the  squares  constructed  upon  the 
lines  plus  twice  the    rectangle   having   the   lines   for   its 
dimensions. 

16,  The  square  constructed  upon  the  difference  of  two 
lines  is  equivalent  to  the  sum  of  the  squares  constructed 
upon  the  lines  minus  twice  the  rectangle  having  the  lines 
for  its  dimensions. 

17,  The  difference  between  the  squares  constructed  upon 
two  lines  is  equivalent  to  the  rectangle  having  the  sum  and 
the  difference  of  the  two  lines  for  its  dimensions. 


104  GEOMETRY. 


18,  If  p  denote  half  the  perimeter  of  a  triangle,  p  the 
radius  of  the  inscribed  circle,  prove  that 

Area  of  a  triangle  =  pp. 

19,  If  a,  b,  c  denote  the  sides  of  a  triangle,  r  the  radius 
of  the  circumscribed  circle,  prove  that 

Area  of  a  triangle  = 

4r 

20,  If  a,  b,  c  denote  the  sides  of  a  triangle,  and  p  = 
\  (a  -f  b  +  c),  prove  that 

Area  of  a  triangle  =  Vp  (p  —  a)(p  —  b)  (p  —  c). 

21,  Of  all  triangles  which  can  be  formed  with  two  given 
lines  for  sides,  that  is  the  maximum  in  which  the  given  lines 
form  a  right  angle. 

22,  Of  all  the  triangles  which  can  be  formed  by  the  two 
sides  of  an  angle  and  lines  drawn  through  a  fixed  point 
in  the  bisector  of  the  angle,  the  isosceles  triangle  is  the 
minimum. 


A  B 

Fig.  88.  Fig. 

23,  Of  all  the  triangles  which  can  be  constructed  upon  a 
given  line  as  base,  and  which  have  a  given  angle  opposite 
this  base,  the  isosceles  triangle  is  the  maximum. 

Construct  upon  the  given  base  a  segment  capable  of  containing 
the  given  angle. 

24,  Of  all  equivalent  triangles  standing  upon  the  same 
base,  the  isosceles  triangle  has  the  minimum  perimeter. 

Of  all   equivalent   triangles,  which   has   the   minimum 
perimeter? 


EQUIVALENT    FIGURES.  105 

Analysis.  (Fig.  88.)  Let  ABC  be  an  isosceles  triangle,  and  ADB 
any  other  triangle,  equal  in  area  and  having  the  same  base  AB. 
Then  CD  is  II  to  AB.  Draw  BE  A.  to  CD,  produce  A  0  to  meet  BE 
in  E,  join  DE,  and  show  that  AC  +  CB<AD  +  DB. 

25.  Of  all  triangles  upon  the  same  base,  and  having  equal 
perimeters,  the  isosceles  triangle  is  the  maximum. 

Analysis.  Let  AB  (Fig.  89)  be  the  given  base,  ABC  the  isosceles 
triangle,  and  ABD  any  other  triangle  having  the  same  perimeter. 
Take  AE=2AC,  draw  CF\\  AB,  and  show  that  D  must  be  between 
CFand  JJ5. 

§  20.   NUMERICAL  EXERCISES. 

Find  the  area  of  a  square,  given  : 

1,  One  side  160  yards.  3,    Perimeter  25  feet. 

2.  One  side  1.5m.  4,    Diagonal  36  yards. 

5,  Find  in  yards  the  side  of  a  square  field  which  con- 
tains 10  acres. 

6,  A  square  field  contains  15f  acres  ;  find  the  length  of 
the  fence  which  encloses  it. 

7,  A  square  plot  of  ground,  the  side  of  which  is  127  yards, 
has  a  path  1  yard  wide  round  the  inside  of  its  perimeter. 
What  will  be  the  expense  of  gravelling  the  path  at  12  cents 
per  square  yard  ? 

8,  What  will  it  cost,  at  18  cents  per  square  foot,  to  tile 
a  walk  2  feet  6  inches  wide  round  the  outside  of  a  square 
court  which  is  30  yards  long? 

'  9,  A  square  field  210  yards  long  has  a  path  round  the 
inside  of  its  perimeter  which  occupies  just  one-seventh  of 
the  whole  field.  Find  the  width  of  the  path. 

Find  the  area  of  a  rectangle,  given  : 

10.    Length  7  feet  6  inches,  and  breadth  6  feet  7  inches. 


106  GEOMETRY. 


11,  Length  210  yards,  and  perimeter  760  yards. 

12,  Length  108  yards,  and  diagonal  135  yards. 

13,  -A  street  li  miles  long  contains  5  acres.    How  wide  is 
the  street? 

14,  How  many  yards  of  paper  27  inches  wide  will  be 
required  to  paper  a  room  18  feet  long,  12  feet  wide,  and  11 
feet  high? 

15,  The  perimeter  of  a  rectangle  is  72  feet,  and  its  length 
is  equal  to  twice  its  width ;  find  its  area. 

16,  A  chain  80  feet  long  encloses  a  rectangle  15  feet 
wide.     How  much  more  surface  would  it  enclose   if  the 
figure  were  a  square  ? 

17,  The  perimeter  of  a  square,  and  also  of  a  rectangle 
whose  length  is  four  times  its  breadth,  is  100  yards.    What 
is  the  difference  in  their  areas  ? 

18,  A  rectangle  whose  length  is  25m  is  equivalent  to  a 
square  whose  side  is  15m.     Which  has  the  greater  perime- 
ter, and  by  how  much  ? 

Find  the  dimensions  of  a  rectangle,  given  : 

19,  Area  216  square  feet,  perimeter  60  feet. 

20,  Area  600  square  feet,  difference  of  sides  10  feet. 

21,  Area  60  square  feet,  diagonal  13  feet. 

22,  Area  756  square  feet,  sides  are  as  7  :  3. 

23,  Area  192  square  feet,  diagonal  %  of  one  side. 

24,  Find  the  area  of  a  rectangle  if  the  diagonal  is  75m, 
and  the  sides  are  as  3  :  4. 

25,  A  rectangle  contains  2400  square  feet,  and  the  length 
is  20  feet  more  than  the  breadth.     Find  its  dimensions. 


EQUIVALENT    FIGURES.  107 

26,  The  dimensions  of  a  rectangle  are  36  feet  by  20  feet. 
If  the  length  is  diminished  by  6  feet,  how  much  must  be 
added  to  the  breadth,  in  order  that  the  area  may  remain 
the  same? 

27,  How  many  tiles  9  inches  long  and  4  inches  wide  will 
be  required  to  pave  a  path  8  feet  wide  surrounding  a  rec- 
tangular court  60  feet  long  and  36  feet  wide  ? 

28,  A  carpenter  requires  3£  square  feet  of  plank,  and  he 
has  a  plank  1£  feet  wide  from  which  to  cut  it  off.     Find 
the  length  of  the  piece  which  he  must  cut  off. 

29,  A  wall  round  a  square  field  at  $0.96  per  yard  ,cost 
$537.60.     What  would  it  have  cost  if  the  field  had  been  in 
the  shape  of  a  rectangle  with  a  length  of  196  yards  ? 

30,  A  man  bought  a  corner  lot  in  the  shape  of  a  rectan- 
gle 121  yards  long  and  100  yards  wide,  at  $5000  per  acre. 
He  laid  out  two  streets  directly  through  the  middle  of  the 
lot,  each  street  being  15  yards  wide  and  parallel  to  two 
opposite  sides  of  the  lot.     The  cost  of  making  the  streets 
was  $2000.     He  then  sold  the  remainder  of  the  land  at  20 
cents  per  square  foot.    Did  he  gain  or  lose,  and  how  much  ? 

31,  The  perimeters  of  two  rectangular  lots  are  102  yards 
and  108  yards,  respectively.     The  first  lot  is  eight-ninths 
as  wide  as  it  is  long,  and  the  second  lot  is  twice  as  long  as 
it  is  wide.    The  value  per  square  foot  of  each  lot  is  $1.00. 
What  is  the  difference  in  the  value  of  the  lots  ? 

Find  the  area  of  a  rhombus,  given  : 

32,  Perimeter  600  feet,  altitude  75  feet. 

33,  Perimeter  40m,  shorter  diagonal  12m. 

34,  The  two  diagonals  18  feet  and  12  feet,  respectively. 

35,  Sum  of  diagonals  12  feet,  their  ratio  3  :  5. 


108  GEOMETRY. 


36,  A  rhombus  contains  100  square  feet,  and  the  shorter 
diagonal  is  10  feet.    Find  the  length  of  the  other  diagonal. 

37,  A  rhombus  and  a  square  have  equal   perimeters. 
Which  has  the  greater  area  ?     What  is  the  ratio  of  their 
areas  if  the  altitude  of  the  rhombus  is  equal  to  one-half  that 
of  the  square  ? 

38,  What  ratio  has  the  area  of  a  rhombus  whose  acute 
angle  is  60°,  to  that  of  a  square  having  an  equal  perimeter  ? 

39,  A  rhombus  and  a  rectangle  have  equal  bases  and 
equal  areas.    Find  their  perimeters  if  one  side  of  the  rhom- 
bus .is  5  feet,  and  the  altitude  of  the  rectangle  is  3£  feet. 

40,  The  area  of  a  rhombus  is  60qm,  and  the  shorter  diag- 
onal is  equal  to  one  side  of  the  rhombus.     Find  the  perim- 
eter. 

41,  The  diagonals  of  a  rhombus  are  90  yards  and  120 
yards,  respectively.     Find  the  length  of  one  side,  and  also 
the  distance  between  the  parallel  sides. 

42,  Find  the  area  of  a  parallelogram  if  the  base  is  40 
feet  6  inches,  and  the  altitude  is  28  feet  9  inches. 

43,  If  two  parallelograms  have  equal  areas,  and  the  base 
of  one  is  three  times  that  of  the  other,  what  is  the  ratio  of 
their  altitudes  ? 

44,  What  is  the  ratio  of  the  areas  of  two  parallelograms 
of  given  bases  and  altitudes  ? 

45,  What  is  the  ratio  of  the  areas  of  two  triangles  of 
given  bases  and  altitudes? 

46,  If  a,  a'  denote  the  bases,  and  A,  h'  the  altitudes, 
respectively,  of  two  equivalent  triangles,  state,  in  the  form 
of  a  proportion,  the  relation  of  the  four  quantities  a,  a',  h,  h1. 


EQUIVALENT    FIGURES.  109 

47,  A  parallelogram  and  a  triangle  have  equal  altitudes, 
but  the  base  of  the  parallelogram  is  equal  to  one-half  that 
of  the  triangle.     Compare  their  areas. 

Find  the  area  of  a  triangle,  given  : 

48,  Base  64  miles,  altitude  14  miles. 

49,  Rt.  A,  legs  6  feet  and  10  feet. 

50,  Rt.  isos.  A,  one  leg  18  feet. 

51,  Rt.  isos.  A,  hypotenuse  20  feet. 

52,  Rt.  A,  one  leg  5  feet,  hypotenuse  13  feet. 

53,  Rt.  A,  one  leg  3  inches,  opposite  angle  30°. 
*  54,  Isos.  A,  base  36  feet,  one  leg  30  feet. 

55,  Isos.  A,  base  2  feet,  angle  at  base  60°. 

56,  Equilat.  A,  one  side  40  feet. 

57,  Equilat.  A,  altitude  24  feet. 

58,  The  altitudes  of  two  triangles  are  equal,  and  their 
bases  are  12  feet  and  16  feet,  respectively.     What  is  the 
ratio  of  their  areas  ? 

59,  The  altitudes  of  two  triangles  are  equal,  and  their 
bases  are  20  feet  and  30  feet,  respectively.     What  is  the 
base  of  a  triangle  equivalent  to  their  sum,  and  having  an 
altitude  one-fourth  as  great? 

60,  A  house  is  40  feet  long,  30  feet  wide,  25  feet  high  to 
the  roof,  and  35  feet  high  to  the  ridgepole.    Find  the  num- 
ber of  square  feet  in  the  entire  exterior  surface. 

61,  What  must  be  the  length  of  the  hypotenuse  of  a 
right  triangle  in  order  that  its  area  may  be  500qm  ? 

62,  Find  the  area  of  a  right  triangle  if  the  perimeter  is 
60  feet,  and  its  sides  are  as  3  :  4  :  5. 


110  GEOMETRY. 


63,  Find  the  area  of  a  right  triangle  if  its  sides  are  as 
3:4:5,  and  the  altitude  upon  the  hypotenuse  is  12  feet. 

64,  The  legs  of  a  right  triangle  are  30  feet  and  40  feet. 
Find   the   areas  of  the  parts  into  which  the   triangle  is 
divided  by  a  perpendicular  drawn  from  the  vertex  of  the 
right  angle  to  the  hypotenuse. 

65,  Find  the  area  of  a  right  triangle  if  one  leg  is  15m, 
and  the  altitude  upon  the  hypotenuse  is  8m. 

66,  The  area  of  a  right  triangle  is  300qm  and  the  hypote- 
nuse is  equal  to  20m.     Find  the  legs. 

67,  The  area  of  a  triangle  is  875  square  feet.     Find  its 
base  and  its  altitude  if  they  are  as  14  :  5. 

68,  ABC  is  a  triangle,  and  AD  the  perpendicular  from 
A  upon  BC.     If  AD  =  13  feet,  and  the  lengths  of  the  per- 
pendiculars from  D  to  AB  and  AC  are  5  feet  and  lOf  feet, 
respectively,  find  the  area  of  the  triangle. 

69,  Find  the  area  of  a  triangle  if  the  three  sides  are  104 
feet,  111  feet,  and  175  feet,  respectively. 

70,  How  many  square  feet  of  carpet  are  required  to 
cover  a  triangular  floor  whose  sides  measure,  respectively, 
26  feet,  35  feet,  and  51  feet? 

71,  The  two  legs  of  a  right  triangle  are  1  foot  and  2  feet. 
Find  the  radius  of  the  inscribed  circle. 

72,  Given  the  sides  a,  b,  c  of  a  triangle ;  find  the  radii 
r,  p  of  the  circumscribed  and  inscribed  circles. 

73,  The  three  sides  of  a  triangle  are :  AB  100  feet,  BC 
89  feet,  AC 21  feet.     Find  the  length  of  the  perpendicular 
from  C  to  AB. 

74,  Find  the  area  of  a  triangle  if  the  perimeter  is  14m 
and  the  radius  of  the  inscribed  circle  is  1.07m. 


EQUIVALENT    FIGURES.  Ill 

Find  the  area  of  a  trapezoid,  given  : 

75,  Bases  50  feet  and  34  feet,  altitude  25  feet. 

76,  Median  25  feet,  altitude  12  feet. 

77,  The  area  of  a  trapezoid  is  700  square  feet,  the  bases 
are  30  feet  and  40  feet,  respectively.     Find  the  distance 
between  the  bases. 

78,  A  trapezoid  contains  240  square  feet,  and  its  altitude 
is  16  feet.     Find  the  two  bases  :  (i.)  if  one  is  3  feet  longer 
than  the  other ;  (ii.)  if  they  are  in  the  ratio  2  :  3. 

79,  The  value  of  a  field  in  the  shape  of  a  trapezoid  is 
$5800.     The  bases  are  200  yards  and  119  yards,  and  the 
distance  between  them  is  110  yards.     Find  the  value  per 
acre. 

80,  The  bases  of  a  trapezoid  are  32  feet  and  20  feet. 
Each  of  the  other  sides  is  equal  to  10  feet.     Find  the  area 
of  the  trapezoid. 

81,  Find  the  area  of  a  trapezoid  if  the  altitude  is  equal 
to  the  median,  the  difference  of  the  bases  is  1  foot,  and  the 
greater  base  is  equal  to  the  hypotenuse  of  a  right  triangle 
whose  legs  are  the  smaller  base  and  the  altitude. 

82,  A  lot  of  land  has  the  shape  of  a  trapezoid.    Its  bases 
are  100m  and  40m.     Each  of  the  other  sides  is  50m.     Find 
(i.)  the  area  in  ars  of  the  trapezoid ;  (ii.)  the  area  of  the 
triangle  formed  by  producing  the  equal  sides  to  their  in- 
tersection. 

83,  Find  the  area  of  a  trapezoid  considered  as  the  sum 
of  a  triangle  and  a  parallelogram. 

84,  Find  the  area  of  a  trapezoid  considered  as  the  differ- 
ence of  two  triangles. 


112 


GEOMETRY. 


85,  A  lot  of  land  has  the  shape  of  a  trapezium.     One 
diagonal  is  108  feet,  and  perpendiculars  upon  it  from  the 
opposite  vertices  are  55  feet  3  inches,  and  60  feet  9  inches, 
respectively.     What  will  the  lot  cost  at  60  cents  per  square 
yard? 

86,  ABGD  is  a  trapezium,  having  AB  =  87  feet,  BO 
=  119  feet,  CD  =  41  feet,  DA  =  169  feet,  AC  =  200  feet. 

Find  the  area. 

87,  Find  the  area  of  a  tangent  quadrilateral  whose  perim- 
eter is  equal  to  400  feet,  the  radius  of  the.  circle  being  25 
feet. 


88,  Find  the  area  of  the  polygon  ABODE  (Fig.  90)  if 
perpendiculars  are  dropped  from  the  vertices  to  a  line  L 
in   the   plane  of  the  figure,  and  the  following  lines  are 
measured : 

AF=  400  feet ;  DI  =  680  feet ;  GH=  250  feet ; 
BH=  200  feet ;  EG  =  700  feet ;  HI  =  200  feet ; 
GK=  380  feet ;  FG  =  150  feet ;  IK  =  220  feet. 

89,  What  is  the  side  of  a  square  equivalent  to  a  rectan- 
gle 200  feet  long  and  32  feet  wide  ? 

90,  The  dimensions  of  a  triangle  are :  base,  360  feet ; 
altitude,    240   feet.     Find   the  altitude  of  an   equivalent 
parallelogram  if  its  base  is  270  feet. 


EQUIVALENT    FIGURES.  113 

91.  A  triangle  whose  base  is  50  feet  is  equivalent  to  a 
rectangle  40  feet  long  and  25  feet  wide.    What  is  the  height 
of  the  triangle  ? 

92.  Upon  the  diagonal  of  a  rectangle  24  feet  long  and  10 
feet  wide  a  triangle  is  constructed   equal  in  area  to  the 
rectangle.     Find  its  altitude. 

93.  A  triangle  whose  base  is  15  feet  and  altitude  8  feet 
is  transformed  into   a  rhombus  whose  altitude  is  6  feet. 
Find  the  perimeter  of  this  rhombus. 

94.  If  a  triangle  whose  base  is  22  feet  and  altitude  15 
feet  is  transformed  into  a  rhombus  whose  longer  diagonal 
is  12  feet,  what  is  the  length  of  the  other  diagonal  ? 

95.  The  bases  of  a  trapezoid  are  16  feet  and  20  feet,  the 
altitude  is  12  feet.     What  is  the  base  of  an  equivalent  rec- 
tangle, having  an  equal  altitude  ? 

96.  The  length  of  a  rectangle  is  60  feet,  the  breadth  40 
feet.     If  the  length  is  reduced  5  feet,  how  much  must  be 
added  to  the  breadth  in  order  that  the  area  may  remain 
the  same  ? 

97.  The  legs  of  a  right  triangle  are  each  18  feet  long. 
Find  the  side  of  an  equivalent  square. 

98.  Find  the  side  of  a  square  equivalent  to  an  equilateral 
triangle  whose  side  is  12  feet. 

99.  Find  the  side  of  an  equilateral  triangle  equivalent 
to  a  square  whose  side  is  12  feet. 

100.  Find  the  side  of  an  equilateral  triangle  equivalent 
to  the  triangle  whose  sides  are  12  feet,  16  feet,  and  18  feet. 

101.  Find  the  side  of  a  square  equivalent  to  a  quadri- 
lateral whose  sides  are  6  feet,  8  feet,  10  feet,  and  12  feet, 
and  one  of  whose  diagonals  is  14  feet. 


114  GEOMETRY. 


102,  Find  the  side  of  a  rhombus  composed  of  two  equi- 
lateral triangles,  and  equivalent  to  another  rhombus  whose 
diagonals  are  12  feet  and  16  feet. 

103,  The  perimeter  of  an  isosceles  triangle,  in  which  the 
base  is  half  the  length  of  one  leg,  is  120  feet.     Find  the 
side  of  a  square  equal  in  area  to  the  triangle. 

104,  The  sides  of  a  triangular  field  are  65  yards,  119 
yards,   and   138  yards,  respectively.     Find  the  side  of  a 
square  field  containing  the  same  amount  of  land. 

105,  The  perimeter  of  a  polygon  circumscribed  about  a 
circle  is  320  feet ;  the  radius  of  the  circle  is  40  feet.     Find 
the  side  of  the  square  equivalent  to  this  polygon. 

106,  A  triangular  lot  of  land  containing  600  square  feet, 
and  having  a  base  of  75  feet,  is  to  be  divided  into  two*  parts 
containing  240  square  feet  and  360  square  feet,  respectively. 
If  the  line  of  division  starts  from  the  opposite  vertex,  how 
will  it  divide  the  base  ? 

107,  A  triangle  containing  80  square  feet  is  divided  by  a 
line  drawn  through  one  vertex  into  two  parts,  one  of  which 
contains  16  square  feet  more  than  the  other.    If  the  base  is 
20  feet,  into  what  two  parts  is  it  divided  by  the  line  of 
division  ? 

108,  A  line  of  division  is  drawn  between  two  sides  of  a 
triangle,  dividing  it  into  a  triangle  and   a  quadrilateral. 
What  parts  are  these  two  figures,  respectively,  of  the  entire 
triangle,  if  the  line  of  division  cuts  off  the  following  parts 
of  the  two  sides,  reckoned  from  the  intersection  of  the  sides? 
(i.)  i  and  i ;    (ii.)  f  and  } ;    (iii.)  |  and  £ ;    (iv.)  £  and  | ; 

(v.)   I  and  -  ;  (vi.)  -  and  1 
m         n  n          n 


EQUIVALENT    FIGURES.  115 

109,  What  parts  of  two  sides  of  a  triangle,  reckoned  from 
their  intersections,  must  be  taken,  in  order  that  the  portion 
of  the  triangle  contained  between  the  two  parts  may  be 
two-fifths  of  the  entire  triangle  ? 

110,  In  a  triangle  ABC,  AB  =  20  feet,  AC  =  64  feet. 
Through  a  point  D  in  AB,  8  feet  from  A,  a  line  DE  is  to 
be  drawn,  meeting  AC  in  JE,  so  that  the  triangle  ADE 
shall  be  three-eighths  of  the  entire  triangle.     Find  AE. 

111,  How  must  two  lines  be  drawn  through  the  middle 
point  of  one  side  of  a  triangle  in  order  that  the  triangle 
may  be  divided  into  parts  which  shall  be  to  each  other  as 
1:2:3? 

112,  A  triangular  field  ABC  is  to  be  divided  into  three 
parts  which  shall  be  as  2:5:3,  by  lines  starting  from  a 
point  D  in  AB,  90  yards  from  A  and  150  yards  from  B. 
If  E,  J^are  the  points  in  AC,  BCto  which  the  lines  should 
be  run,  find  AE  and  BF'm  terms  of  A C  and  BC,  respec- 
tively. 

Through  a  point  D  in  the  side  AB  of  the  triangle  ABC 
a  line  DE  is  drawn  parallel  to  BG.  Find  the  value  of  AD 
in  terms  of  AB  if : 

113,  DE  divides  triangle  ABC  into  two  equivalent  parts. 

114,  Triangle  ADE=  £  triangle  ABC. 

115,  Trapezoid  BDEC=  \  triangle  ABC. 

116,  Triangle  ADE :  trapezoid  BDEC=  ra  :  n. 

117,  The  base  AB  of  a  triangle  is  40  feet  in  length. 
The  foot  D  of  the  altitude  is  distant  30  feet  from  A  ;  to  find 
a  point  E  in  AB  such  that  the  perpendicular  erected  at  E 
will  divide  the  triangle  into  two  parts  having  the  ratio  3  :  5. 


116  GEOMETRY. 

118,  What  part  of  a  parallelogram  is  the  triangle  cut  off 
by  a  line  drawn  from  one  vertex  to  the  middle  point  of  one 
pf  the  opposite  sides  ? 

119,  The  area  of  a  parallelogram  is  60  square  feet.     In 
what  ratio  must  a  line  through  one  vertex  divide  the  oppo- 
site side,  in  order  that  the  triangle  cut  off  may  contain  25 
square  feet? 

120,  What  part  of  a  parallelogram  is  contained  between 
one-half  of  one  side  and  two-thirds  of  the  opposite  side  ? 

121,  A  parallelogram   is   divided   into   two  equivalent 
trapezoids  by  a  line  which  divides  one  side  in  the  ratio 
1:2.    In  what  ratio  does  this  line  divide  the  opposite  side  ? 

122,  In    the   trapezoid  A  BCD  it  is  required  to  draw 
through  the  point  jE,  in  the  upper  base  DC,  a  line  EF 
which  shall  divide  the  trapezoid  into  two  equivalent  parts. 
Find  AF,  given  AB  =  6  feet,  DC=  4£  feet,  DE=  1|  feet. 

123,  In  the  trapezoid  ABCD  the  bases  are  AB  =  5  feet, 
DC=  3  feet.    To  determine  a  point  X'm  the  diagonal  DB, 
such  that  a  line  through  X,  and  parallel  to  AD,  will  divide 
the  trapezoid  into  two  parts  having  the  ratio  2 :  3. 

124,  A  trapezoid  whose  bases  are  12  feet  and  7  feet, 
respectively,  and  whose  altitude  is  denoted  by  h,  is  divided 
into  two  equivalent  parts  by  a  line  parallel  to  the  bases. 
Find  the  length  of  this  line  and  the  altitude  of  the  upper 
part. 

125,  In  a  trapezoid,  given  the  two  bases  a,  b,  and  the 
altitude  h.     The  legs  are   divided  each  into  three  equal 
parts  by  lines  parallel  to  the  bases.     Find,  in  terms  of  a,  b, 
and  h,  the  areas  of  the  three  parts  into  which  the  trapezoid 
is  divided. 


EQUIVALENT   FIGURES.  117 

126.  In  two  similar  polygons  two  homologous  sides  are 
15  feet  and  25  feet.     The  area  of  the  first  polygon  is  450 
square  feet ;  what  is  the  area  of  the  other  polygon  ? 

127.  The  base  of  a  triangle  is  32  feet,  and  its  altitude  20 
feet.     What  is  the  area  of  the  triangle  formed  by  drawing 
a  line  parallel  to  the  base  and  5  feet  from  the  vertex? 

128.  The  altitude  of  a  triangle  is  6  feet.     What  is  the 
homologous  altitude  of  a  similar  triangle  100  times  as  large  ? 

129.  If  one  side  of  a  triangle  is  a,  what  is  the  value  of 
the  homologous  side  of  a  similar  triangle  one-half  as  large  ? 

130.  The  side  of  a  square  is  4  feet.     Find  the  side  of  a 
square  which  shall  be  to  the  given  square  as  3  : 5. 

131.  The  side  of  a  square  is  4  feet.     Find  the  side  of  a 
square  3^-  times  as  large. 

132.  The  diagonals  of  two  squares  are  12  feet  and    18 
feet.     Find  the  diagonal   of  a  square  equivalent  to  their 
sum. 

133.  The  sides  of  two  equilateral  triangles  are  3  feet  and 
4  feet.    What  is  the  side  of  the  equilateral  triangle  equiva- 
lent to  their  sum  ? 

134.  The  homologous  sides  of  two  similar  hexagons  are 
33  feet  and  56  feet.     Find  the  homologous  side  of  a  similar 
hexagon  equivalent  to  their  sum. 

135.  What  is  the  ratio  of  the  square  constructed  upon 
the  diagonal  of  a  given  square  to  the  given  square  ? 

136.  What  is  the  ratio  of  the  square  constructed  upon  a 
given  line,  and  the  isosceles  right  triangle  constructed  upon 
this  line  as  (i.)  hypotenuse  ;  (ii.)  leg  ? 

137.  If  the  side  of  one  equilateral  triangle  is  equal  to 
the  altitude  of  another,  find  the  ratio  of  their  areas. 


118  GEOMETRY. 


138,  The  sides  of  a  triangle  are  10  feet,  17  feet,  and  21 
feet.     Find  the  areas  of  the  two  parts  into  which  the  tri- 
angle is  divided  by  the  bisector  of  the  angle  formed  by  the 
first  two  sides. 

139,  In  a  trapezoid  one  base  is  10  feet,  the  altitude  is  4 
feet,  and  the  area  is  32  square  feet.     Find  the  length  of  a 
line  drawn  between  the  legs  parallel  to  the  given  base  and 
distant  1  foot  from  it. 

140,  Find   the   sides  of  an  isosceles  trapezoid   circum- 
scribed about  a  circle,  knowing  that  the  sum  of  the  bases 
is  2  a,  and  that  the  area  is  equal  to  that  of  a  square  whose 
side  is  b.     Examine  the  case  in  which  a  is  equal  to  b. 

Find  the  area  F  of  a  triangle,  given  : 

141,  Rt,  A,  a,  c.  145,  Equilat.  A,  a. 

142,  Rt.  A,  a,  h.  146.  Equilat,  A,  h. 

143,  Rt.  A,  qt  h.  147,  Isos.  A,  a,  c. 

144,  Rt.  A,  jo,  q.  148,  Isos.  A,  a,  h. 

149,    IPOS.  A,  h,  ha. 

150,  In  an  equilateral  triangle  given  F;  find  a  and  h. 

151,  If  the  altitude  h  of  a  triangle  is  increased  by  the 
length  m,  by  how  much  must  the  base  a  be  diminished  in 
order  that  the  area  may  remain  the  same  ? 

152,  Find  the  change  in  the  area  of  a  triangle  of  base  a 
and  altitude  b  in  the  following  cases : 

(i.)  if  a  and  h  are  increased  by  m  and  n,  respectively  ; 

(ii.)  if  a  and  h  are  diminished  by  m  and  n,  respectively  ; 

(iii.)  if  a  is  increased  by  m,  and  h  diminished  by  n  ; 

(iv.)  if  a  is  diminished  by  m,  and  h  increased  by  n. 

153,  Find  the  area  .Fof  a  square,  given  (i.)  the  sum  s  of 
a  diagonal  and  a  side ;  (ii.)  the  difference  d  of  a  diagonal 
and  a  side. 


EQUIVALENT    FIGURES.  119 


154,  The  distances  of  a  point  within  a  rectangle  from 
tha  sides  a,  b  are  m,  n,  respectively.     Through  this  point 
parallels  to  the  sides  are  drawn.     Find  the  areas  of  the 
parts  into  which  the  rectangle  is  divided. 

155,  Find  the  diagonal  /  of  a  parallelogram,  given  the 
sides  a,  b,  and  the  other  diagonal  g. 

156,  The  sides  of  a  rectangle  are  a,  b.     Find  the  length 
of  a  line  which  shall  have  the  same  ratio  to  the  perimeter 
of  the  rectangle  as  its  square  has  to  the  area  of  the  rect- 
angle. 

157,  Find  the  sides  of  a  rectangle,  given  the  diagonal/ 
and  the  area  F. 

158,  Find  the  diagonals  of  a  rhombus,  given  the  side  a 
and  the  area  F. 


159,  In  the  trapezoid  A  BCD,  given  the  bases  AB  =  a, 
CD  =  c.  Find  the  lengths  x,  y  of  lines  parallel  to  the 
bases  and  dividing  the  trapezoid  into  parts  which,  counted 
from  CD,  are  to  one  another  as  m  :  n  :  p. 


§  21.   PROBLEMS. 

For  the  notation  in  the  triangle  ABC,  see  page  44 ;   also,  I  and  k 
are  used  to  denote  given  lengths,  </>  a  given  angle. 

To  construct  a  triangle  equivalent  to : 

1,  The  sum  of  two  given  triangles  with  equal  altitudes. 

2,  The  difference  of  two  given  triangles  with  equal  alti- 
tudes. 

3,  The  sum  of  two  given  triangles  with  equal  bases. 

4,  The  difference  of  two  given  triangles  with  equal  bases. 

5,  n  times  as  large  as  a  given  triangle. 


120  GEOMETRY. 


To  transform  the  triangle  AJBCinto  : 

6,  A  triangle  with  c  unchanged  and  I  instead  of  a. 

7,  A  triangle  with  c  unchanged  and  <f>  instead  of  a. 

8,  A  triangle  with  c  unchanged  and  I  instead  of  ha. 

9,  A  right  triangle  with  c  for  one  leg. 

10,  A  right  triangle  with  c  for  hypotenuse. 
Hi    An  isosceles  triangle  with  c  for  base. 

12,  An  isosceles  triangle  with  c  for  one  leg. 

13,  A  rectangle  with  c  for  one  side. 

14,  A  parallelogram  with  c  and  a  unchanged. 

15,  A  parallelogram  with  a  and  b  for  diagonals. 

16,  A  rhombus  with  a  for  one  diagonal. 

17,  To  transform  the  triangle  ABC  into  a  triangle  with 
I  instead  of  <?,  and  a  unchanged. 

Analysis.    Upon  AB  (produced  if  necessary)  take  AD  =  I,  draw 
BE  II  to  CD  and  meeting  AC  (produced  if  necessary)  in  E\    then 


To  transform  the  triangle  ABC  into  : 

18,  A  triangle  with  I  instead  of  c,  and  <£  instead  of  a. 

19,  A  triangle  with  /  instead  of  c,  and  k  instead  of  b. 

20,  A  triangle  with  I  instead  of  <?,  and  k  instead  of  ha. 

21,  A  triangle  with  I  instead  of  c,  and  k  instead  of  m. 

22,  A  right  triangle  with  I  for  one  leg. 

23,  A  right  triangle  with  I  for  hypotenuse. 

24,  An  isosceles  triangle  with  I  for  the  base. 

25,  An  isosceles  triangle  with  I  for  one  leg. 

26,  A  triangle  with  I  instead  of  <?,  and  <£  instead  of  y. 

27,  A  triangle  with  I  instead  of  r,  and  </>  instead  of  y. 

28,  A  triangle  with  I  instead  of  c,  and  k  instead  of  r. 


EQUIVALENT    FIGURES.  121 

29,  To  transform  the  triangle  ABC  into  a  triangle  with 
I  instead  of  h,  and  a  unchanged. 

Analysis.  Draw  AD  _L  to  AS  and  equal  to  I,  DE  II  to  AS,  and 
meeting  AC  (produced  if  necessary)  in  E,  CF\\  to  EB  and  meeting 
AB  (produced  if  necessary)  in  F\  then  A  AEF*>&  ABO. 

To  transform  the  triangle  ABC  into: 

30,  A  triangle  with  I  instead  of  h,  and  <£  instead  of  a. 

31,  A  triangle  with  I  instead  of  h,  and  Ic  instead  of  in. 

32,  A  triangle  with  I  instead  of  h,  and  k  instead  of  ha. 

33,  A  right  triangle  with  I  for  altitude  upon  hypote- 
nuse. 

34,  An  isosceles  triangle  with  I  for  altitude  upon  the 


35,  A  triangle  with  I  instead  of  h,  and  <f>  instead  of  y. 

36,  A  triangle  with  I  instead  of  h,  and  Jc  instead  of  r. 

37,  A  triangle  with  I  instead  of  h,  and  <f>  instead  of  Z  am. 

38,  A  triangle  with  one  vertex  in  a  given  point. 

To  construct  a  triangle  equivalent  to  : 

39,  The  sum  of  two  given  triangles. 

40,  The  difference  of  two  given  triangles. 

To  transform  the  triangle  ABO  into  : 

41,  A  triangle  with  base  and  altitude  equal. 

42,  A  triangle  with  base  and  altitude  equal,  and  /  in- 
stead of  a. 

43,  A  triangle  with  base  and  altitude  -equal,  and  I  in- 
stead of  ra. 

44,  A  triangle  with  base  and  altitude  equal,  and  </>  in- 
stead of  y. 


122  GEOMETRY. 


45,    To  transform  the  triangle  ABC  into  a  triangle  sim- 
ilar to  a  given  triangle  PQR. 


A  E  F  B       P  Q 

Fig.  91. 

Analysis.  Let  AFG  (Fig.  91)  be  the  required  triangle  constructed 
with  the  side  AF  in  the  line  AB.  Draw  CD  II  to  AB  and  meeting 
AG  in  D,  and  DE  II  to  GF;  then,  by  similar  triangles, 

A  AED  -.  A  AFG  =  AE2  -.  AF2  • 
or,  since  A  AFG  —  A  ABC, 

A  AED  •.  A  ABC  =  AE2  -.  AF2. 

Since  the  triangles  AED  and  ABC  have  equal  altitudes, 
A  AED  :  A  ABC=  AE-..AB. 
.-.AE2-.  AF2  =  AE:AB, 
and  A  F2  =  AB  x  AE. 

That  is,  AF  is  the  mean  proportional  between  the  two  known 
lines  AB  and  AE. 

To  transform  the  triangle  ABC  into  : 

46,  A  right  triangle  with  <f>  for  one  of  the  acute  angles. 

47,  An  isosceles  triangle  with  <£  for  the  angle  at  the 
vertex. 

48,  An  equilateral  triangle. 

49,  A  triangle  with  one  angle  unchanged  and  the  oppo- 
site side  parallel  to  a  given  line. 

50,  To  transform  the  triangle  ABC  into  a  square. 

51,  To   transform  the  triangle  ABO  into  a  trapezoid 
with  c  for  one  base  and  for  one  adjacent  angle,  and  <f>  for 
the  other  adjacent  angle. 


EQUIVALENT    FIGURES.  123 

To  transform  the  triangle  A BC  into  : 

52,  A  triangle  with  a  given  perimeter. 

53,  A  right  triangle  with  a  given  perimeter. 

54,  A  right  triangle  with  I  for  radius  of  inscribed  circle. 

55,  To  transform  a  right  triangle  into  an  isosceles  trian- 
gle having  an  angle  in  common  with  the  right  triangle. 

To  transform  a  parallelogram  into : 

56,  A  parallelogram  having  a  given  side  I. 

57,  A  parallelogram  having  a  given  angle  <j>. 

58,  A  parallelogram  having  a  given  altitude  I. 

To  transform  a  square  into  : 

59,  A  right  triangle.  60,    An  isosceles  triangle. 

61,  An  equilateral  triangle. 

62,  A  rectangle  having  a  given  side. 

63,  A  rectangle  having  a  given  perimeter. 

64,  A  rectangle  having  a  given  difference  of  its  sides. 

65,  A  rectangle  having  a  given  diagonal. 

To  transform  a  rectangle  into : 

66,  A  square.  67.    An  equilateral  triangle. 

68,  A  rectangle  having  a  given  side. 

69,  A  rectangle  having  a  given  perimeter. 

70,  A  rectangle  having  a  given  difference  of  its  sides. 

71.  A  rectangle  having  a  given  diagonal. 

To  construct  a  parallelogram  equivalent  to  the : 

72.  Sum  of  two  given  parallelograms  of  equal  altitudes. 

73.  Difference  of  two  given  parallelograms  of  equal  alti- 
tudes. 


124  GEOMETRY. 


74,  Sums  of  two  given  parallelograms  of  equal  bases. 

75,  Difference  of  two  given  parallelograms  of  equal  bases. 

76,  Sum  of  two  given  parallelograms. 

77,  Difference  of  two  given  parallelograms. 

To  transform  a  parallelogram  into  : 

78,  A  triangle.  80,    An  isosceles  triangle. 

79,  A  right  triangle.  81.    An  equilateral  triangle. 

82,  A  square. 

83,  A  rhombus  having  for  diagonal  one  side  of  the  paral- 
lelogram. 

84,  A  rhombus  having  a  given  diagonal. 

85,  A  rhombus  having  a  given  side. 

86,  A  rhombus  having  a  given  altitude. 

87,  A  parallelogram  having  given  diagonals. 

88,  A  parallelogram  having  a  given  side  and  a  given 
diagonal. 

89,  To  transform  a  rhombus  into  a  square. 

90,  To  inscribe  in  a  given  circle  a  rectangle  equivalent 
to  a  given  square. 

To  transform  a  trapezoid  into : 

91,  A  triangle.  92,    A  square. 

93,  A  parallelogram  having  for  base  one  base  of  the 
trapezoid. 

94,  An  isosceles  trapezoid. 

To  transform  a  trapezium  into : 

95,  A  triangle. 

96,  An  isosceles  triangle  having  a  given  base. 

97,  A  parallelogram. 


EQUIVALENT    FIGURES. 


125 


98,  A  trapezoid    with    one  side  and  the  two  adjacent 
angles  unchanged. 

99,  To  transform  a  given  polygon   M  into  a  polygon 
similar  to  a  given  polygon  N. 

Hint.    First  transform  M  and  N  into  squares. 

100,  Given  an  angle  BA  C  and  a  point  P  in  AB ;  through 
P  to  draw  a  line  cutting  AC  in  Q  so  that  the  triangle  APQ 
-shall  be  equivalent  to  a  given  square. 

101,  Through  a  given  point  P  to  draw  a  line  cutting  the 
sides  of  a  given  angle  BA  C  in  X  and  Y  so  that  the  triangle 
AXY  shall  be  equivalent  to  a  given  triangle  DEF. 


B    .,-'   X        H 
Fig.  92. 


Transform  the  given  A  DEF  (Fig.  92)  into  the  A  ABC  having 
the  given  angle  for  one  of  its  angles.  Draw  PR  II  to  AB,  and  trans- 
form the  A  ABC  into  the  A  ARII,  having  R  for  vertex,  and  the  base 
AS"  in  the  line  AB.  Then  transform  the  A  ARH  inio  the  parallel- 
ogram AR8M,  having  the  side  AR  and  the  angle  A  in  common  with 
the  A  ARH.  Since  A  AXY  must  be  equal  to  this  parallelogram, 
A  PQS=  A  PRY  +  A  QMX.  Since  these  three  triangles  are  sim- 
ilar, 

PR2  ^A  PRY  AndMX2^&MQY. 
PS*     APQtf'          PS2      APQS' 

whence  PZ?  +  HX*  =  PS*. 

Therefore,  the  point  X  is  found  by  constructing  the  rt.  A  MNX, 
in  which  MNis  _L  to  AB  and  equal  to  PR,  and  NX  =  PS. 


126  GEOMETRY. 


102,  To   construct  a  square  which  shall  be  to  a  given 
square  as  5  :  3. 

To  construct  a  square  equivalent  to : 

103,  One-half  of  a  given  square. 

104,  One-third  of  a  given  square. 

105,  Five-eighths  of  a  given  square. 

106,  Nine-sevenths  of  a  given  square. 

107,  Three-fifths  of  a  given  pentagon. 

108,  The  sum  of  two  given  equilateral  triangles. 

109,  The  sum  of  a  given  triangle  and  a  given  rectangle. 

110,  To  construct  upon  a  given  line  as  base  a  triangle 
equivalent  to  the    difference    between    a   given    rectangle 
and  a  given  trapezoid. 

111,  To  construct  a  polygon  equivalent  to  the  sum  of 
two  given  polygons  M,  Nt  and  similar  to  a  third  given 
polygon  R. 

112,  To  construct  a  polygon  equivalent  to  the  difference 
of  two  given  polygons  M,  JV,  and  similar  to  a  third  poly- 
gon JR. 

To  divide  into  n  equivalent  parts  (e.g.,  n  =  5): 

113,  A  triangle.  115,    A  trapezoid. 

114,  A  parallelogram.  116.    A  trapezium, 
Analysis  for  Ex.  116.    Divide  a  diagonal  into  n  equal  parts,  and 

join  the  points  of  division  to  the  opposite  vertices. 

To  divide  into  two  parts  having  the  ratio  m:n\  (e.g., 
ra=2,  n=3): 

117,  A  triangle.  119,    A  trapezoid. 

118,  A  parallelogram.  120,    A  trapezium. 

To  divide  into  three  parts  which  shall  be  to  one  another 
as  m:n\p\  (e.g.,  m  =  1,  n  =  2, p  =  3)  : 

121,    A  triangle.  122,    A  parallelogram. 


EQUIVALENT    FIGURES.  127 

123,    A  trapezoid.  124,    A  trapezium. 

125,  To  divide  the  triangle  ABC  into  five  equivalent 
parts  by  a  broken  line  passing  from  C  to  c,  from  c  to  a, 
from  a  to  c,  and  from  c  to  a. 

To  divide  the  triangle  ABC,  by  lines  through  a  given  point 
P  in  ^4.5,  into  : 

126,  Two  equivalent  parts. 

127,  Three  equivalent  parts. 

128,  Five  equivalent  parts. 

129,  Two  parts  in  the  ratio  3  :  4. 

Analysis  for  Ex.  126.    If  PQ  is  the  required  line,  If  the  middle 
point  of  AB,  then  A  PCQ  =  A  PCM-  .:  MQ  is  II  to  CP. 

130,  To  cut  from  the  triangle  ABC  a  triangle  equivalent 
to  a  given  triangle  by  a  line  of  division  drawn  from  a  point 
Pin  AC. 

Hint.    Transform  the  given  triangle  into  another  with  CP  for  one 
side  and  ACB  for  an  adjacent  angle. 

131,  To  find  a  point  within  a  triangle  such  that  the  lines 
joining  this  point  to  the  vertices  shall  divide  the  triangle 
into  three  equivalent  parts. 

Hint.    Draw  the  medians. 

132,  To  find  a  point  within  a  triangle  such  that  the  lines 
joining  this  point  to  the  vertices  shall  divide  the  triangle 
into  parts  proportional  to  the  numbers  m,  n,p ;  (e.g.,  m  =  1, 
n=2,p  =  B). 

133,  To  divide  a  triangle  into  five  equivalent  parts  by 
lines  drawn  from  a  given  point  within  the  triangle. 

To  divide  a  triangle  by  lines  parallel  to  one  side  into : 

134,  Two  equivalent  parts. 

135,  Three  equivalent  parts. 

136,  Two  parts  in  the  ratio  3  :  5. 


128  GEOMETRY. 


137,  Parts  proportional  to  1,  2,  3. 

138,  To  divide  a  triangle  into  two  equivalent  parts  by  a 
line  perpendicular  to  one  side. 

139,  To  divide  a  triangle  into  three  equivalent  parts  by 
lines  parallel  to  one  of  the  medians. 

140,  To  divide  a  triangle  into  four  equivalent  parts  by 
lines  parallel  to  one  of  the  bisectors. 

141,  To  divide  a  triangle  into  two  equivalent  parts  by  a 
line  drawn  through  a  given  point. 

142,  To  divide  a  triangle  into  two  parts  in  the  ratio  2 :  5 
by  a  line  drawn  through  a  given  point. 

To  divide  by  lines  drawn  from  one  vertex : 

143,  A  parallelogram  into  three  equivalent  parts. 

144,  A  parallelogram  into  four  equivalent  parts. 

145,  A  parallelogram  into  five  equivalent  parts. 

146,  A  parallelogram  into  two  parts  in  the  ratio  3  :  4. 

147,  A  trapezoid  into  two  equivalent  parts. 

148,  A  trapezoid  into  three  equivalent  parts. 

149,  A  trapezium  into  two  equivalent  parts. 

150,  A  trapezium  into  two  parts  in  the  ratio  1 :  2. 

151,  An  octagon  into  two  equivalent  parts. 

To  divide  a  parallelogram  by  lines  drawn  from  a  given 
point  in  one  of  the  sides  into  : 

152,  Two  equivalent  parts.     153,    Five  equivalent  parts. 

154,  Two  parts  in  the  ratio  3:5. 

To  divide  a  parallelogram  into  two  equivalent  parts  by 
a  line : 

155,  Drawn  through  a  given  point  P. 

156,  Parallel  to  a  given  line  L. 


EQUIVALENT    FIGURES.  129 


To  divide  a  trapezoid  into  two  equivalent  parts  by  a  line : 

157.  Parallel  to  the  bases. 

158,  Perpendicular  to  the  bases. 

159,  Parallel  to  one  of  the  legs. 

160,  Drawn  through  a  given  point  in  one  of  the  bases. 

161.  Drawn  through  a  given  point  P. 

162,  Parallel  to  a  given  line  L. 

163.  To  divide  a  trapezoid  by  lines  parallel  to  the  bases 
into  three  parts  proportional  to  the  numbers  1,  2,  3. 

164.  To  divide  a  trapezium  into  two  equivalent  parts  by 
a  line  drawn  through  a  given  point  in  one  of  the  sides. 

'    165.    To  cut  from  a  given  polygon  a  triangle  equivalent 
to  a  given  square. 

To  divide  a  hexagon  by  a  line  drawn  through  a  given 
point  in  one  of  the  sides  into  : 

166,  Two  equivalent  parts. 

167,  Two  parts  in  the  ratio  2  :  3. 

168,  To  inscribe  in  a  given  circle  a  rectangle  of  given  area. 

169,  To  inscribe  in  a  given  triangle  a  rectangle  of  given 
area. 

170,  To  inscribe  in  a  given  parallelogram  a  rhombus  of 
given  area. 


.  130  GEOMETRY. 


CHAPTER    V. 

REGULAR    FIGURES. 


§  22.    THEOREMS. 

1,  The  side  of  a  circumscribed  equilateral  triangle  is  equal 
to  twice  the   side    of  the    inscribed    equilateral   triangle. 
What  is  the  ratio  of  their  areas  ? 

2,  The  area  of  a  circumscribed  square  is  equal  to  twice 
the  area  of  the  inscribed  square.    What  is  the  ratio  of  their 
sides  ? 

3,  The  apothem  of  an  inscribed  equilateral  triangle  is 
equal  to  one-half  the  side  of  the  inscribed  regular  hexagon. 

4,  The  apothem  of  an  inscribed  regular  hexagon  is  equal 
to  one-half  the  side  of  the  inscribed  equilateral  triangle. 

5,  An  inscribed  regular  hexagon  is  twice  as  large  as  the 
inscribed  eqiiilateral  triangle. 

6,  A  regular  inscribed  hexagon  is  one-half  as  large  as 
the  circumscribed  equilateral  triangle. 

7,  The  area  of  a  regular  dodecagon  is  equal  to  three 
times  the  square  of  its  radius. 

8,  Upon  the  six  sides  of  a  regular  hexagon  squares  are 
constructed  outwardly.      Prove  that  the  exterior  vertices 
of  these  squares  are  the  vertices  of  a  regular  dodecagon. 

9,  In  a  regular  pentagon  all  the  diagonals  are  drawn. 
Prove  that  another  regular  pentagon  is  thereby  formed. 


REGULAR    FIGURES.  131 


10,  The  apothem  of  an  inscribed   regular  pentagon  is 
equal  to  one-half  the  sum  of  the  radius  of  the  circle  and 
the  side  of  the  inscribed  regular  decagon. 

11,  The  side  of  an  inscribed  regular  pentagon  is  equal  to 
the  hypotenuse  of  a  right  triangle  of  which  the  legs  are  the 
radius  of  the  circle  and  the  side  of  the  inscribed  regular 
decagon. 

12,  The  radius  of  an  inscribed  regular  polygon  is  the 
mean  proportional  between  the  apothem  and  the  radius  of 
the  similar  circumscribed  regular  polygon. 

13,  The  area  of  a  circular  ring  is  equal  to  that  of  a  circle 
whose  diameter  is  a  chord  of  the  outer  circle  and  a  tangent 
to  the  inner  circle. 

14,  The  alternate  vertices  of  a  regular  hexagon  are  joined 
by  straight  lines.     Prove  that  another  regular  hexagon  is 
formed.     Find  the  ratio  of  the  areas  of  the  two  hexagons. 

15,  If  upon  the  legs  of  a   right   triangle  semi-circum- 
ferences are  described   outwardly,  the   sum  of  the   areas 
contained  between  these  semi-circumferences  and  the  semi- 
circumference  passing  through  the  three  vertices,  is  equal 
to  the  area  of  the  triangle. 

16,  If  the  diameter  of  a  circle  is  divided  into  two  parts, 
and  upon  these  parts  semi-circumferences  are  described  on 
opposite  sides  of  the  diameter,  these  semi-circumferences  will 
divide  the  circle  into  two  parts  which  have  the  same  ratio 
as  the  two  segments  of  the  diameter. 

17,  If  in  a  circle  two  chords  are  drawn  perpendicular  to 
each  other,  and  upon  the  four  segments  of  these  chords  as 
diameters  circles  are  described,  the  sum  of  the  areas  of  the 
four  circles  will  be  equal  to  the  area  of  the  given  circle. 


132 


GEOMETRY. 


§  23.   NUMERICAL  EXERCISES. 

NOTE.   When  TT  occurs  as  a  factor,  take  IT  =  ty  if  English  units  are 
used,  and  ir  =  3.1416  if  metric  units  are  used. 

In  the  following  twenty-one  exercises : 

r  =  radius  of  circle. 

p  =  apothem  of  inscribed  regular  polygon  of  n  sides. 
a  =  one  side  of  inscribed  regular  polygon  of  n  sides. 
b  =  one  side  of  inscribed  regular  polygon  of  2n  sides. 
c  =  one  side  of  circumscribed  regular  polygon  of  n  sides. 
JF—  area  of  inscribed  regular  polygon  of  n  sides. 
G=  area  of  inscribed  regular  polygon  of  2n  sides. 

In  case  approximate  rational  results  are  desired,  \/2  = 
1.41421,  v/3=  1.73205,  \/5  =  2.23606. 


Given. 

^e^wirec?. 

1, 

2, 

3, 
4, 

5, 
6, 

rt  a. 

r,  a. 

r,  b. 
r,  a. 

r,  n  =  3. 
r,  n  =  6. 

/          ,,? 

P  =^r2-       =£\/4r2-a2. 

J_V2r(r-^-^} 

V62(4r2-62) 
r 

2ar 

V4r2-a2 

•  -tVS,  p-|. 

4 

rx/3 
«-*,    p-=j- 

REGULAR   FIGURES. 


13S 


Given. 

.Retired 

7, 

r,  n=12. 

V°     \/3                 rV^+V^ 

2 

8, 
9, 

r,  n  =  4. 

r,  n  =*  8. 

P==—  • 

P              2 

10, 

r,n=  1Q. 

*-tf-'*      p-|VioT^ 

11, 

r,  n  =  5. 

«-|vi?r^.  p=;-(V5+D. 

12, 

a,  n,  p. 

2  ' 

13, 

a,  n,  r. 

^=  — 

14, 

a,  n,  r. 

ilftt 

15, 

a  or  r,  n  =  3. 

^     a2V3      3r2V3 

4              4 

16, 

a  or  r,  ?i  =  6. 

r     3a2V3      3r2\/3 

2               2 

17, 

a  or  r,  w=  12. 

^3r2  =  3a2(2  +  V3). 

18, 

a  or  r,  w  =  4. 

^^a2  =  2r2. 

19, 

a  or  r,  n  =  8. 

^2,2V2  =  2a2(l+V2). 

20, 
21, 

r,  n  =  5. 
r,  n  =  10. 

~~8~ 

~T 

134  GEOMETRY. 


Hints.    In  all  cases,  r,  p,  and  £  a  are  the  sides  of  a  right  triangle, 

r  being  the  hypotenuse. In  Ex.  9,  a  may  be  found  either  by  a 

direct  investigation  or  by  making  use  of  the  result  in  Ex.  2.  — 
Ex.  10.  Since  the  radius  is  divided  in  extreme  and  mean  ratio,  we 

have  r :  a  =  a :  r  —  a.     From  this  proportion  find  a  in  terms  of  r. 

In  comparing  the  sides  of  the  pentagon  and  the  decagon,  the  expres- 
sions A/5  +  1  and  V5  —  1  occur.     It  should  be  noted  with  care  that 


\/5  +  1  =  V5  +  2V5  +  1  =  V6  +  2V5, 


and  that          VB  -  1  =  V5  -  2 \/5  +  1  =  V 6  -  2\/5. 

Ex.  14.   The  polygon  of  2n  sides  is  composed  of  2n  equal  isosceles 
triangles.     If  in  each  triangle  the  radius  of  the  circle  is  taken  as  the 

base,  the  altitude  will  be  equal  to  Ja. In  Exs.  17,  19,  21,  the 

result  obtained  in  Ex.  14  may  be  immediately  applied. 

22,  Apply  the  general  formula  obtained  in  Ex.  4  to  the 
circumscribed    equilateral    triangle,    square,    and    regular 
hexagon. 

23,  Find  the  area  of  a  regular  polygon  of  24  sides  in- 
scribed in  a  circle  whose  radius  is  1  foot. 

24,  Find  the  area  of  a  regular  dodecagon  circumscribed 
about  a  circle  whose  radius  is  r. 

25,  Compare  the  areas  of  a  circumscribed  regular  hexa- 
gon and  an  equilateral  triangle  inscribed  in  the  same  circle. 

26,  If  the  alternate  vertices  of  a  regular  hexagon  are 
joined  by  drawing  six  diagonals,  prove  that  another  regular 
hexagon  is  formed,  and  find  the  ratio  of  its  area  to  that  of 
the  given  hexagon. 

27,  Find  the  perimeter  of  a  regular  pentagon  inscribed 
in  a  circle  whose  radius  is  12  feet. 

28,  Find  one  side  of  a  regular  octagon  circumscribed 
about  a  circle  whose  radius  is  r. 


REGULAR   FIGURES.  135 


29.  Given  the  side  c  of  a  circumscribed  regular  decagon, 
find  the  side  of  a  regular  decagon  inscribed  in  the  same 
circle. 

30.  Two  regular  octagons  contain,  respectively,  54  square 
feet  and  62  square  feet.     Find,  approximately,  the  side  of 
a  third  regular  octagon  equal  in  area  to  the  sum  of  the  two 
given  octagons. 

31.  The  sides  of  three  regular  octagons  are  3  feet,  4  feet, 
and  12  feet,  respectively.     Find  the  side  of  a  regular  octa- 
gon equal  in  area  to  the  sum  of  the  areas  of  the  three  given 
octagons. 

32.  The  length  of  each  side  of  a  park  in  the  shape  of  a 
regular  decagon  is  80  yards.     Find  the  area  of  the  park. 

33.  What  will  it  cost,  at  $1.80  per  yard,  to  build  a  wall 
around  a  lot  of  land  in  the  shape  of  a  regular  hexagon, 
containing  1039.24  square  yards? 

34.  The  perimeter  of  a  regular  hexagon  is  480  feet,  and 
that  of  a  regular  octagon  is  the  same.    Which  is  the  greater 
in  area,  and  by  how  much  ? 

35.  Find  the  different  ways  in  which  regular  polygons 
may  be  employed  for  paving. 

36.  Find  the  circumference  of  a  circle  whose  diameter  is 
7  inches. 

37.  Find  the  diameter  of  a  circle  whose  circumference  is 
12  feet  10  inches. 


38.  The  diameter  of  a  carriage-  wheel  is  3£  feet. 
many  revolutions  does  it  make  in  traversing  one-fourth  of 
a  mile  ? 

39.  If  a  carriage-wheel  makes  220  revolutions  in  travel- 
ling half  a  mile,  find  its  diameter. 


136  GEOMETRY. 


40,  What  is  the  width  of  the  ring  between  two  concen- 
tric circumferences  whose  lengths  are  440  feet  and  330  feet? 

41,  In  raising  water  from  the  bottom  of  a  well,  by  means 
of  a  wheel,  it  is  found  that  the  wheel  whose  diameter  is 
2  feet  4  inches  makes  30  revolutions  in  raising  the  bucket. 
Find  the  depth  of  the  well. 

42,  Find  the  length  of  an  arc  of  36°  in  a  circle  whose 
diameter  is  35  inches. 

43,  Find  the  angle  subtended  at  the  centre  by  an  arc  6 
feet  5  inches  long,  if  the  radius,  of  the  circle  is  8  feet  2 
inches. 

44,  The  radius  of  a  circle  is  7  inches,  and  the  length  of 
an  arc  is  the  same.    Find  the  angle  subtended  at  the  centre 
by  the  arc. 

45,  Find  the  angle  subtended  at  the  centre  by  an  arc 
whose  length  is  equal  to  the,  radius  of  the  circle. 

46,  The  radius  of  a  circle  is  5  feet  3  inches.    What  is  the 
perimeter  of  a  sector  the  angle  of  which  is  45°  ? 

47,  If  the  angle  subtended  at  the  centre  by  an  arc  5  feet 
6  inches  long  is  72°,  tind  the  radius  of  the  circle. 

48,  Find  the  length  of  an  arc  of  11°  15'  in  a  circle  whose 
radius  is  84  feet. 

49,  The  radius  of  a  circle  is  14  feet.    What  is  the  length 
of  the  arc  subtended  by  one  side  of  the  inscribed  regular 
dodecagon  ? 

50,  Taking  the  length  of  a  meridian  of  the   earth   as 
40,000,000m,  find  the  length  of  an  arc  of  1". 

51,  Find  in  kilometers  the  radius  of  the  earth,  assuming 
the  circumference  to  be  40,000,000m. 


REGULAR    FIGURES.  137 


52,  Two  arcs  have  the  same  angular  measure,  but  the 
length  of  one  is  twice  that  of  the  other.    Compare  the  radii 
of  the  arcs. 

53,  Two  arcs   have   the   same    length.     Their   angular 
measures  are  20°  and  30°,  respectively.     If  the  radius  of 
the  first  arc  is  6  feet,  find  the  radius  of  the  second  arc. 

54,  Find  the  area  of  a  circle  whose  diameter  is  1  foot 
2  inches. 

55,  Find  the  area  of  a  circle  whose   circumference   is 
2  miles. 

56,  Find  the  diameter  of  a  circle  whose  area  is  6  acres. 

57,  Find  the  circumference  of  a  circle  whose  area  is 
2  acres  176  square  yards. 

58,  Find  the  area  of  a  semicircle  the  radius  of  which  is 
14  feet. 

59,  The  diameter  of  a  circle  is  56  feet.     Find  the  side  of 
a  square  equal  in  area  to  the  circle. 

60,  The  area  of  a  square  is  196  square  feet.     Find  the 
area  of  the  circle  inscribed  in  the  square. 

61,  A  circular  fish-pond  which  covers  an  area  of  3  acres 
880  square  yards,  is  surrounded  by  a  walk  3  yards  wide. 
Find  the  cost  of  gravelling  the  walk,  at  8  cents  per  square 
yard. 

62,  What  must  be  the  width  of  a  walk  around  a  circular 
plot  of  land  containing  2  acres  2794  square  yards,  in  order 
that  the  walk  may  contain  exactly  one-fourth  of  an  acre  ? 

63,  The  diameter   of  a  circular  grass-plot  is  28  feet. 
Find  the  diameter  of  another  circular  plot  which  is  just 
twice  as  large. 


138  GEOMETRY. 


64,  One  side  of  a  square  piece  of  wood  is  5  feet  10  inches 
long ;  out  of  it  is  cut  the  largest  possible  circle.    How  many 
square  inches  of  wood  are  cut  away  ? 

65,  Out  of- -a,  circular  piece  of  wood  whose   diameter  is 
3  feet  4  inches  is  cut  the  largest  possible  square.     Find  the 
length  of  its  side. 

66,  Find  the  area  between  two  concentric  circles  if  their 
circumferences  are  460  feet  and  352  feet. 

67,  In  a  square  room  whose  side  measures  17  feet  6 
inches  is  to  be  made  a  circular  bath  with  its  circumference 
touching  the  walls  of  the  room.     Find   the   area   of  the 
bath. 

68,  The  perimeters  of  a  circle,  a  square,  and  an  equilat- 
eral triangle  are  each  equal  to   132  feet.     Compare  the 
areas  of  the  three  figures. 

69,  The  radius  of  a  circle  is  3  feet.     What  is  the  radius 
of  a  circle  25  times  as  large  ? 

70,  Find  the  radius  of  a  circle  equal  in  area  to  the  dif- 
ference between  the  areas  of  two  given  circles  with  the 
radii  r,  r'. 

71,  What  is  the  ratio  of  the  area  of  a  circle  to  that  of  the 
inscribed  equilateral  triangle  ? 

72,  What  is  the  ratio  of  the  area  of  the  circle  inscribed 
in  an  equilateral  triangle  to  that  of  the  circle  circumscribed 
about  the  same  triangle  ? 

73,  The  sum  of  the  areas  of  a  circle  and  the  inscribed 
equilateral  triangle  is  3qm.  .  Find  the  area  of  each  figure. 

74,  What  will  it  cost  to  pave  a  circular  court  30  feet  in 
diameter,  at  54  cents  per  square  foot,  leaving  in  the  centre 
a  hexagonal  space,  each  side  of  which  measures  2  feet  ? 


REGULAR   FIGURES.  139 


75,  A  circle  18  feet  in  diameter  is  divided  into  three 
equivalent  parts  by  two  concentric  circumferences.  Find 
the  radii  of  these  circumferences. 

.  76.  Calculate  the  expense  of  making  a  moat  round  a 
circular  island,  at  60  cents  per  square  yard,  the  diameter 
of  the  island  being  525  feet,  and  the  breadth  of  the  moat 
being  21  feet  6  inches. 

77.  The  chord  of  half  an  arc  is  17  feet,  and  the  height 
of  the  arc,  7  feet.     Find  the  diameter  of  the  circle. 

78.  The  chord  of  half  an  arc  is  12  feet,  and  the  radius 
of  the  circle  is  18  feet.     Find  the  height  of  the  arc. 

79.  The  height  of  an  arc  is  9  inches,  and  the  diameter 
of  the  circle  is  125  feet.     Find  the  chord  of  half  the  arc. 

80.  The  chord  of  an  arc  is  24  inches,  and  the  height  of 
the  arc  is  9  inches.     Find  the  diameter  of  the  circle. 

81.  The  span  (chord)  of  a  bridge  in  the  form  of  a  cir- 
cular arc  is  96  feet,  and  its  height  above  the  stone  piers  is 
12  feet.     Find  the  radius  of  the  circle. 

82.  The  height  of  the  arch  of  a  bridge,  the  form  of  which 
is  the  arc  of  a  circle,  is  24  feet,  and  the  radius  of  the  circle 
is  312  feet.     Find  the  span  of  the  arch. 

83.  The  radius  of  a  circle  is  12  feet.     The  chords  which 
subtend  two  arcs  AB,  BO  are  6  feet  and  9  feet,  respec- 
tively.    Find  the  chord  AC. 

84.  The  lengths  of  two  chords  drawn   from   the  same 
point  in  the  circumference  of  a  circle  to  the  extremities  of 
a  diameter  are  6  feet  and  8  feet.     Find  the  area  of  the 
circle. 

85.  The  chord  of  an  arc  is  32  inches,  and  the  radius  of 
the  circle  is  34  inches.     Find  the  length  of  the  arc. 


140  GEOMETRY. 


86,  The   diameter  of  a   circle   is    106   feet.     Find   the 
lengths  of  the  two  arcs  into  which  a  chord  90  feet  long 
will  divide  the  circumference. 

87,  The  span  of  a  bridge,  the  form  of  which  is  the  arc 
of  a  circle,  is  200  feet,  and  the  height  of  the  crown  above 
the  stone  piers  is  42  feet.     Find  the  length  of  the  arch. 

88,  Find  the  area  of  a  sector  if  the  radius  of  the  circle  is 
56  feet,  and  the  angle  at  the  centre  is  22z°. 

89,  The  area  of  a  sector  is  385  square  feet,  ancl  the  angle 
of  the  sector  is  36°.     Find  the  radius  of  the  circle  and  the 
perimeter  of  the  sector. 

90,  The  chord  of.  an  arc  is  40  feet,  and  the  radius  of  the 
circle  is  21  feet.     Find  the  area  of  the  sector. 

91,  Find  the  area  of  a  segment  if  the  chord  of  the  arc  is 
56  feet,  and  the  radius  of  the  circle  is  35  feet. 

92,  The  chord  of  a  segment  is  40  feet,  and  its  height  is 
9  feet.     Find  its  area. 

93,  A  room  20  feet  long  and  15  feet  wide  has  a  recess  at 
one  end  in  the  shape  of  the  segment  of  a  circle,  the  chord 
of  the  recess  being  15  feet,  and  its  greatest  width  4  feet. 
Find  the  area  of  the  whole  room. 

94,  Find  how  many  square  feet  of  brick  are  required  in 
blocking  up  one  of  the  arches  of  a  railway  viaduct,  if  the 
span  of  the  arch  is  60  feet,  height  above  the  piers  is  20 
feet,  distance  from  the  ground  to  the  spring  of  the  arch  is 
20  feet. 

Compute  in  terms  of  the  radius  r  of  the  circle  the  area 
of  the  segment  subtended  by  one  side  of  the  inscribed : 

95,  Equilateral  triangle.         98,    Regular  octagon. 

96,  Regular  hexagon.  99,    Regular  pentagon. 

97,  Square.  100,    Regular  decagon. 


REGULAR    FIGURES.  141 


101,  Find  the  area  of  a  circle  in  which  a  chord  3  feet 
long  subtends  an  arc  of  120°. 

102,  Find  the  area  of  a  circle  if  the  area  of  the  inscribed 
regular  hexagon  is  10  square  feet. 

103,  Find  the  area  of  a  segment  whose  arc  is  300°,  the 
radius  of  the  circle  being  1  foot. 

104,  Find  the  ratio  of  the  two  segments  into  which  a 
circle  is  divided  by  a  perpendicular  erected  at  the  middle 
point  of  one  of  the  radii. 

105,  The  radius  of  a  circle  is  7m.     Find  the  areas  of  the 
two  segments  into  which  the  circle  is  divided  by  a  chord 
equal  to  the  radius. 

106,  Two  tangents  drawn  from  a  point  exterior  to  a  circle 
whose  radius  is  1.3™  intercept  upon  the  circumference  an 
arc  whose  length  is  0.57m.     Required  the  angle  formed  by 
the  two  tangents. 

107,  The  areas  of  two  concentric  circles  are  as  5  to  8. 
The  area  of  that  part  of  the  ring  which  is  contained  between 
two  radii  making  the  angle  45°  is  300  square  feet.     Find 
the  radii  of  the  two  circles. 

108,  What  is  the  altitude  of  a  rectangle  equivalent  to  a 
sector  whose  radius  is  15  feet,  if  the  base  of  the  rectangle 
is  equal  to  the  arc  of  the  sector  ? 

109,  The  radii  of  two  concentric  circles  are  3  feet  and 
5  feet.     What  is  the  altitude  of  a  trapezoid  equivalent  to 
the  ring  and  having  for  bases  the  lengths  of  the  two  cir- 
cumferences ? 

110,  Find  the  radius  of  a  circle  if  its  area  is  doubled  by 
increasing  the  radius  1  foot. 

111,  The  sides  of  a  rectangle  inscribed  in  a  circle  are 
a  b.     Find  the  area  of  the  circle  in  terms  of  a  and  b. 


142 


GEOMETRY. 


112,  The  radius  of  a  circle  is  r.     Upon  the  chord  of  a 
quadrant  as  diameter  a  semi-circumference  is  described  out- 
wardly.    Find  the  area  of  the  surface  contained  between 
the  semi-circumference  and  the  arc  of  the  quadrant. 

113,  The  radius  of  a  circle  is  r.     Through  a  point  exte- 
rior to  the  circle  two  tangents  are  drawn  making  an  angle 
of  60°.     Find  the  area  of  the  figure  bounded  by  the  tan- 
gents and  the  intercepted  arc. 

114,  Three  equal  circles  are  described,  each  touching  the 
two  others.     If  the  common  radius  is  r,  find  the  area  con- 
tained between  the  circles. 

115,  In  a  quadrant  whose  radius  is  r  a  circle  is  inscribed. 
Find  (i.)  the  area  of  this  circle,  (ii.)  the  area  comprised 
between  the  circle  and  the  arc  of  the  quadrant. 

116,  Upon  each  side  of  a  square  as  a  diameter  semi- 
circumferences  are  described  within    the   square    forming 
four  leaves  {Fig.  93).     If  a  side  of  the  square  is  a,  find  the 
sum  of  the  areas  of  the  four  leaves. 

H 


Fig.  93.  Fig.  94. 

117,  In  a  circle  whose  radius  is  r,  two  parallel  chords 
are  drawn  on  the  same  side  of  the  centre,  one,  AB,  equal  to 
the  side  of  the  inscribed  regular  hexagon,  the  other,  CD, 
equal  to  the  side  of  the  inscribed  equilateral  triangle.  Find 
the  area  of  the  portion  of  the  circle  contained  between  the 
two  chords. 


REGULAR    FIGURES. 


143 


This  question  may  be  solved  algebraically  by  computing  the  dif- 
ference between  the  segments  CHD  and  AHB  (Fig.  94).  But  the 
following  geometrical  solution  is  shorter.  Let  E,  F,  H  be  the  middle 
points  of  AB,  CD,  and  arc  AB,  respectively,  and  let  G  be  the  inter- 
section of  CD  and  OB.  The  rt.  &  EOB  and  FOD  are  equal  (why  ?). 
By  subtracting  the  common  part  FOG  and  adding  GBD  it  becomes 
evident  that  the  figures  FEBD  and  BOD  are  equivalent.  But  sector 
BOD  is  equivalent  to  sector  HOB,  because  the  A  BOD  and  HOB 
are' each  equal  to  30°.  Therefore,  area  A  BCD,  which  is  double 
that  of  FEBD,  is  equivalent  to  sector  AOB,  which  is  double  the 
sector  HOB.  Now  sector  AOB  =  £irr2,  .-.  area  ABCD  =  £irr2. 

118.  Given  p,  Pthe  perimeters  of  regular  polygons  of 
n  .sides  inscribed  in  and  circumscribed  about  a  given  circle, 
findjt/,  F  the  perimeters  of  regular  polygons  of  2n  sides 
inscribed  in  and  circumscribed  about  the  same  circle. 


Solution.  Let  (Fig.  95)  AB  be  one  side  of  the  polygon  whose 
perimeter  is  p,  0  the  middle  point  of  the  arc  AB.  Join  AC,  draw 
through  C  a  tangent  meeting  OA  and  OB  produced  in  E  and  F  re- 
spectively ;  also  draw  tangents  through  A  and  B,  meeting  EF  in 
G  and  _ff  respectively  ;  join  OG,  and  let  OG  cut  A  C  in  M. 


Then 


AC  =  side  of  polygon  whose  perimeter  is  p1 ; 
EF  =  side  of  polygon  whose  perimeter  is  P; 
GH=  side  of  polygon  whose  perimeter  is  P. 


144  GEOMETRY. 


Now,  p:P=OA:OE  =  00:  OE; 

00  :  OE  =  CG  :  OE. 
.'.p:  P  =  CG:  GE. 
By  the  theory  of  proportions, 

p  :  P  +  p  =  CG  :  OE. 

But,  CG  =  —  ,  CE=-£. 

4n  2n 

Substituting  these  values  of  CG  and  OE,  and  solving  for  P,  we  obtain 


P+P 

Secondly,  the  &  AD(7and  CGMare  similar. 
:.AC-.AD  =  CG:  CM. 

Now,         AC=^L,   AD  =  2--    CG  =  ^--  CM=^- 
2n  2n  4n  4n 

Substituting  these  values,  and  solving  for  p',  we  obtain 
p'  =  VP^. 

119,    Show  how  to  compute  approximately  the  value  of  v 
by  means  of  the  formulas  deduced  in  the  last  exercise. 

If  C  denote  the  circumference  of  a  circle,  and  D  its  diameter,  then 

C       C  C 

by  definition,  v  =  —  —  —  .     If  we  make  r  =  1,  then  *  =  —  .     Now, 

D     2r  2 

by  means  of  the  formulas  of  the  last  exercise,  we  can  obtain  the  value 
of  C  to  as  close  a  degree  of  approximation  as  we  please.  For  exam- 
ple, beginning  with  the  square,  we  havep  =  4  V2,  P=  8  ;  whence  we 
find  the  perimeters  p1,  P'  of  the  inscribed  and  circumscribed  regular 
octagons,  and  then  those  of  regular  polygons  of  16  sides,  etc.  For 
polygons  of  256  sides,  p'  =  6.28303,  P1  =  6.28350.  Since  the  value  of 
C  lies  between  those  of  p'  and  P',  therefore  the  value  0=  6.283  is 
correct  to  three  decimal  places  ;  and  the  same  is  true  of  the  value 


This  method  of  determining  the  value  of  IT  is  the  one  commonly 
given  in  text-books,  and  is  known  as  the  Method  of  Perimeters. 


REGULAR   FIGURES. 


145 


120.  Given  the  radius  r  and  the  apothem  p  of  an  in- 
scribed regular  polygon  of  n  sides :  find  the  radius  r'  and 
the  apothem  p'  of  an  isoperimetrical  regular  polygon  of  2  n 


Hints.  Let  AS  (Fig.  96)  be  one  side  of  the  given  polygon,  C  its 
middle  point.  Produce  OC  to  D,  join  AD  and  BD,  and  join  E,  F, 
the  middle  points  of  AD,  BD,  by  a  line  EF,  cutting  OD  in  G.  EF 
=  side  of  required  polygon  (why  ?)  ;  /.  EOF=  angle  at  its  centre 
(why?).  Hence,  OE=r',  OG  =  p'.  The  following  results  are  now 
readily  found  : 


and 


2  2 

It  is  evident,  both  from  the  figure  and  from  these  values  o 
r1,  that  p'  >  p,  and  that  r7  <  r. 

121.    Show  how  to  compute  approximately  the  value  of  TT 
by  the  aid  of  the  formulas  deduced  in  the  last  exercise. 
Fora  side  whose  square  is  1,  we  have  perimeter  •=  4, 


Now,  with  the  aid  of  the  formulas  of  the  last  exercise,  we  can  com- 
pute the  radius  and  the  apothem  of  the  isoperimetrical  regular  octa- 
gon, then  those  of  isoperimetrical  regular  polygons  of  16,  32,  64, 
etc.,  sides.  As  we  increase  the  number  of  sides  the  radius  decreases, 
the  apothem  increases,  while  the  perimeter  remains  constantly  equal 
to  4.  The  radius  and  the  apothem  both  approach,  as  a  common 
limit,  the  radius  of  a  circle  whose  circumference  is  equal  to  4  ;  in  other 
words,  the  value  of  the  radius  of  this  circle  lies  between  the  values 
of  the  radius  and  of  the  apothem  of  the  polygon,  and  hence  can  be 
found  to  any  desired  degree  of  approximation. 


146  GEOMETRY. 


For  a  polygon  of  8192  sides,  pf  =  0.6366196 ,  r'  =0.6366196 : 

hence,  the  value  0.6366196  must  be  the  correct  value  to  seven  places 
of  decimals  of  the  radius  of  a  circle  whose  circumference,  is  equal  to  4. 
Whence,  we  have 


This  method  of  computing  the  value  of  it  is  known  as  the  Method 
of  Isoperimeters. 

§  24.   PROBLEMS. 

1,  Construction  of  a  Kegular  Polygon,  The  parts  of  a  regu- 
lar polygon  are  the  number  of  sides  n,  the  length  of  one 
side  a,  the  radius  r,  and  the  apothem  p.  The  angle  at  the 
centre  and  the  angle  of  the  polygon  are  determined  by  the 
value  of  n,  and  therefore  are  not  independent  parts  of 
the  polygon.  Neither  can  any  two  of  the  three  parts  a,  r,  p, 
have  arbitrary  values,  but  only  those  values  which  will 
make  the  angle  at  the  centre  an  aliquot  part  of  360°. 

A  regular  polygon  is  determined  by  two  independent 
parts,  because  it  is  divisible  into  equal  right  triangles. 
The  sides  of  each  triangle  are  r,  p,  £  a ;  the  acute  angles 
are  the  angle  of  the  polygon  and  half  of  the  angle  at  the 
centre,  both  of  which  are  known  if  n  is  given. 

From  what  has  been  said,  it  follows  that  the  construction 
of  a  regular  polygon  falls  under  one  of  three  cases : 

(i.)  To  construct  a  regular  polygon,  given  r  and  n. 

(ii.)  To  construct  a  regular  polygon,  given  p  and  n. 

(iii.)  To  construct  a  regular  polygon,  given  a  and  n. 

In  each  case,  the  values  of  n  for  which  an  exact  con- 
struction can  be  given  are  limited  to  the  terms  of  the  four 
series  in  No.  231.  Moreover,  large  values  of  n  are  excluded 
on  account  of  the  difficulty  in  effecting  the  construction. 


REGULAR    FIGURES. 


147 


2,  To  find  the  simplest  construction  for  the  sides  of 
an  inscribed  regular  pentagon  and  an  inscribed  regular 
decagon. 


0  D 

Fig.  97. 


Let  BO  (Fig.  97)  be  one  side  of  the  inscribed  regular  decagon. 
Then  (No.  228)  OS2  =  r  (r  -  OS), 


whence,  by  solving,     OB  +  -=  A/r2  +  —  • 

Therefore,  OB  +  -  is  equal  to  the  hypotenuse  of  a  right  triangle 

of  which  the  legs  are  r  and  -  . 
2 

Hence,  to  construct  OB,  draw  0(7  JL  to  AO,  take  OD  =  -;  from 
D  as  centre,  with  Z)(7as  radius,  cut  OA  in  B:  then-O.S  will  be  one 
side  of  the  inscribed  regular  decagon.  Also, 

BC*  =  r*  +  OB* 


=  r2  +  -  (V5  -  I)2 
4 

=  r2  +  -  (6  -  2  V5) 


6r2  _ 


23,  Ex.  10.) 


!-(10-2V5), 
4 


Therefore  (§  23,  Ex.  11),  5(7  is  equal  to  the  side  of  the  inscribed 
regular  pentagon. 


148  GEOMETRY. 


To  inscribe  in  a  given  circle : 

3,  An  equilateral  triangle.  6,  A  regular  octagon. 

4,  A  square.  7,  A  regular  pentagon. 

5,  A  regular  hexagon.  8,  A  regular  decagon. 

To  circumscribe  about  a  given  circle : 
9,    An  equilateral  triangle.      12,    A  regular  octagon. 

10.  A  square.  13,    A  regular  pentagon. 

11,  A  regular  hexagon.  14,    A  regular  decagon. 

To  construct  upon  a  given  line  as  one  side : 

15.  An  equilateral  triangle.     18,    A  regular  octagon. 

16.  A  square.  19,    A  regular  pentagon. 

17.  A  regular  hexagon.  20,    A  regular  decagon. 

21,  To  construct  upon  a  given  line  as  one  side  a  regular 
dodecagon. 

22,  To  construct  upon  a  given  line  as  one  side  a  regular 
polygon  of  15  sides. 

23,  To  construct  an  angle  of  36°. 

24,  To  construct  an  angle  of  9°. 

25,  To  construct  an  angle  of  12°. 

26,  In  a  given  circle,  to  construct  a  mean  proportional 
between  a  chord  of  30°  and  a  chord  of  150°. 

27,  To  transform  a  given  regular  octagon  into  a  square. 

28,  To  construct  a  regular  hexagon,  given  one  of  the 
shorter  diagonals. 

29,  To  construct  a  regular  pentagon,  given  one  of  the 
diagonals. 


REGULAR    FIGURES.  149 


30,    To  inscribe  in  a  given  circle  a  polygon  of  n  sides,  n 
being  any  whole  number. 


Fig.  98. 

The  following  construction  is  found  in  most  cases  to  be  sufficiently 
exact  for  practical  purposes. 

Divide  the  diameter  AB  (Fig.  98)  into  n  equal  parts  (in  the  Figure 
n  =  7).  Draw  the  radius  CD  JL  to  AB,  produce  CB  to  E,  and  CD  to 
F,  making  BE  and  DF  each  equal  to  one  of  the  parts  of  the  diame- 
ter; join  EF,  cutting  the  circle  for  the  first  time  in  G.  Then  the 
line  GH  joining  G  and  the  third  point  of  division  of  AB,  counting 
from  B,  will  be  very  nearly  equal  to  one  side  of  the  inscribed 
polygon  of  n  sides. 

For  n  =  3  and  n  =  4,  this  construction  is  impossible ;  for  n  =  5  it 
is  useless,  on  account  of  its  inaccuracy  ;  but  for  n  >  5  it  gives  a  very 
close  approximation  to  the  exact  value  of  the  side  required. 

31.  To  construct  a  regular  decagon  equivalent  to  a  given 
regular  pentagon. 

32,  To  construct  a  circle  in  which  the  inscribed  regular 
octagon  shall  have  a  perimeter  equal  to  that  of  a  given 
square. 

33.  To  draw  through  a  given  point  a  line  which  shall 
divide  a  given  circumference  into  two  parts  in  the  ratio  of 
3  to  7. 

34,  To  construct  a  circumference  equal  to  the  sum  of  two 
given  circumferences. 


150  GEOMETRY. 


35,  Three  given  circumferences  are  denoted  by  c,  c1,  c"  ; 
to  construct  a  circumference  equal  to  %c  -f  f  c1  —  f-c". 

36,  To  construct  a  circle  equivalent  to  the  sum  of  two 
given  circles. 

37,  To  construct  a  circle  equal  in  area  to  three  times  a 
given  circle. 

38,  To  construct  a  circle  equal  in  area  to  three-fourths 
of  a  given  circle. 

39,  To  divide  a   given  circle  by  a  concentric  circum- 
ference into  two  equivalent  parts. 

40,  To  divide  a  given  circle  by  concentric  circumferences 
into  four  equivalent  parts. 

41,  To   inscribe    four  equal  circles  in  a   given   regular 
octagon  so  that  each  circle  shall  touch  two  other  circles  and 
one  side  of  the  octagon. 

42,  To  inscribe  in  a  given  circle  five  equal  circles  so  that 
each  circle  shall  touch  two  other  circles  and  the  circum- 
ference of  the  given  circle. 


THE    ALGEBRAIC    METHOD.  151 

CHAPTER  VI. 

THE  ALGEBRAIC  METHOD. 


§  25.   CONSTRUCTION  OF  ALGEBRAIC  EXPRESSIONS. 

The  lengths  of  lines,  expressed  in  terms  of  a  common 
unit,  are  called  Linear  Factors,  Abstract  numbers  and  sym- 
bols representing  abstract  numbers  are  called  Coefficients, 

In  the  following  exercises  the  first  letters  of  the  alphabet, 
a,  b,  <?,  etc.,  are  employed  as  linear  factors,  and  the  letters 
m  and  n  denote  coefficients. 

I.    Elementary  Expressions. 

1,  To  construct  x  =  a  -f  b. 

In  other  words,  to  construct  a  line  equal  to  the  sum  of  two  given 
lines.  The  construction  is  obvious. 

2,  To  construct  x  =  a  —  b. 

Draw  OA  =  a,  then  from  A  take  in  the  opposite  direction  AB  =  b; 
OB  represents  the  value  of  a;. 


If  6  >  a,  the  value  of  x  is  negative.  In  this  case,  the  problem  has 
no  meaning  if  absolute  magnitude  alone  is  considered.  But  if  posi- 
tion is  also  taken  into  account,  the  negative  value  of  x  is  represented 
by  a  line  OB  extending  from  0  in  the  opposite  direction  to  that  of  a 
positive  value  of  x,  or  that  of  the  positive  line  OA. 

This  exercise  furnishes  a  simple  illustration  of  the  principle  of 
Descartes,  that  contrary  signs  in  Algebra  correspond  to  opposite  direc- 
tions in  Geometry .  Or,  stated  more  exactly,  if  a  positive  value  of  a 


152  GEOMETRY. 


linear  expression  is  properly  represented  by  a  line  drawn  from  a  cer- 
tain point  in  a  certain  direction,  a  negative  value  will  be  represented 
by  a  line  drawn  from  the  same  point  in  the  opposite  direction. 

3,  To  construct  x  =  a  +  b  —  e-\-d  —  e  +/. 

4,  To  construct  x  —  na. 

5,  To  construct  x  —  -• 

n 

6,  To  construct  x  — 

n 
The  solution  consists  in  dividing  a  in  the  ratio  m :  n. 

7,  To  construct  x  =  ab. 

In  this  case,  x  is  represented  by  the  area  of  a  rectangle  whose  sides 
are  equal  to  a  and  b. 

8,  To  construct  x  =  a2. 

9,  To  construct  x  =  — 

o 

x  is  the  fourth  proportional  to  a,  b,  and  c,  and  constructed  by 
means  of  No.  139  or  162  or  163. 

10,  To  construct  x  =  -• 

o 

x  is  the  third  proportional  to  a  and  b,  and  is  constructed  by  means 
of  No.  157  or  158  or  164. 

11,  To  construct  x  =  -\fab. 

x  is  the  mean  proportional  between  a  and  b,  and  is  constructed  by 
means  of  No.  157  or  158  or  164. 

12,  To  construct  x  =  Va2  +  b*. 

x  is  equal  to  the  hypotenuse  of  a  right  triangle  whose  legs  are  a 
and  b. 

13,  To  construct  x  =  Va2  —  &2. 

x  is  equal  to  one  leg  of  a  right  triangle  in  which  the  hypotenuse 
is  a,  and  the  other  leg  b.  Also,  since  Va2  —  62  =  V(a  +  b)(a  -  b), 
x  is  the  mean  proportional  between  a  +  b  and  a  —  b. 


THE    ALGEBRAIC    METHOD.  155 

14,  To  construct  x  —  a V2. 

x  is  equal  to  the  diagonal  of  a  square  whose  side  is  a. 

15,  To  construct  x  =  -  VS. 

Zi 
x  is  equal  to  the  altitude  of  an  equilateral  triangle  whose  side  is  a. 

16,  To  construct  x  =  a  V5. 

x  is  equal  to  the  hypotenuse  of  a  right  triangle  whose  legs  are  « 
and  2  a. 

17,  To  construct  x  —  a  Vm. 

a;  is  the  mean  proportional  between  a  and  ma. 

18,  To  construct  x  =±=       a6      . 

V^+T2 
SB  is  equal  to  the  altitude  of  a  right  triangle  whose  legs  are  a  and  b. 

19,  To  construct  x  =  Va2  -f  b2  —  ab. 

x  is  the  third  side  of  a  triangle  in  which  the  other  sides  are  a  and  b, 
and  the  angle  included  by  them  is  60°. 


20,  To  construct  x  =  Va2  -f-  b2  +  ab. 

x  is  the  third  side  of  a  triangle  in  which  the  other  sides  are  a  and 
b,  and  the  angle  included  by  them  is  120°. 

II.    Expressions  Reducible  to  Elementary  Expressions. 
To  construct  the  following  expressions  : 

21.  x=  Va2  +  &2  +  c2-foT2.  23,   x  =  aV3. 


22.   x  =  V«2  -  b2  +  c2  -  cP.  24.   ^  = 

Three  different  constructions  are  suggested  by  putting  a  A/19  into 
the  forms  Va  X  19  a, 


and  V(5a)2-(2a)2-a2-a2. 

25,   3?  =  aVl5.  26.   a?= 

27.   *  =  22*£=2aft.x£.  28,   *= 


ejg 


154 


GEOMETRY. 


44,    x  =  A/a4      a2^2  =     «Va 


46,    To  construct  the  roots  of  the  equation  x2  —  ax  -f-  S2=  0. 

It  is  shown  in  Algebra  that  if  a  complete  equation  of  the  second 
degree  is  written  in  the  form  x2  +px  +  q  =  Q,  the  algebraic  sum  of  the 
roots  =  —  _p,  the  product  of  the  roots  =  q2. 


THE    ALGEBRAIC    METHOD.  155 


Therefore,  in  the  present  case,  if  xl  or2  denote  the  roots, 


These  relations  show  that  the  roots,  if  real,  are  both  positive,  since 
their  sum  a  and  product  b2  are  both  positive  ;  and  the  question  is 
reduced  to  the  problem  :  To  construct  two  lines,  given  their  sum  a 
and  their  product  b2. 


A         E  0  B 

Fig.  99. 

Draw  (Fig.  99)  AB  =  a,  describe  upon  AB  as  diameter  a  semi- 
circle, then  draw  AC-L  to  -4.Z?  and  equal  to  b,  CD  II  to  AB  and  meet- 
ing the  circle  in  D,  DE±  to  AB,  meeting  AB  in  E.  The  required 
lines  are  AE  and  £"5.  (No.  157). 

If  b>  -,  or  a<  26,  (7Z>  will  not  meet  the  circle,  and  the  roots  are 
imaginary.  If  a  =  26,  CD  touches  the  circle,  and  the  roots  are  equal 
each  to  -• 

47,  To  construct  the  roots  of  the  equation  #2+a#-f-62=0. 

Since  this  equation  is  obtained  from  the  equation  x2  —  ax+b2  =  0 
by  changing  x  to  —  x,  its  roots  will  be  the  same  as  those  of  x2  —  ax  + 
62  =  0,  with  contrary  signs.  Therefore,  their  absolute  values  may  be 
constructed  exactly  as  in  the  last  exercise. 

48,  To  construct  the  roots  of  the  equation  a?—-cas—&=Q. 

In  this  case  the  roots  have  contrary  signs,  since  their  product  -  62 
is  negative.  Let  xt  denote  the  positive  root,  x2  the  absolute  value  of 
the  negative  root ;  then 


and  the  question  is  reduced  to  the  problem :  To  construct  two  lines, 
given  their  difference  a  and  their  product  b2. 


156  GEOMETRY. 


With  0  (Fig.  100)  as  centre,   and  -  as  radius,  describe  a  circle. 

Through  any  point  A  of  the  circumference  draw  a  tangent  AB  =  b  ; 
then  draw  BO  meeting  the  circumference  in  C  and  D.  The  required 
lines  are  BG  and  BD.  (No.  164.) 


B  A 

Fig.  100. 

49,    To  construct  the  roots  of  the  equation  x*  -\-  ax — bz  =  0. 

This  equation  is  obtained  from  the  equation  x2  —  ax  —  bz  =  0  by 
changing  x  to  —  x ;  therefore  its  roots  are  the  same  as  those  of  xz  —ax 
—  b2  =  0,  with  contrary  signs,  and  their  absolute  values  are  con- 
structed in  the  same  way. 

NOTE.  Every  complete  quadratic  equation  containing  linear  fac- 
tors can  be  put  into  one  of  the  four  forms  discussed  in  Exs.  46-49. 

§  26.   HOMOGENEITY  OF  ALGEBRAIC  EXPRESSIONS. 

The  Degree  of  a  rational  algebraic  term,  containing  linear 
factors,  is  equal  to  the  number  of  linear  factors  in  the  nu- 
merator diminished  by  the  number  of  linear  factors  in  the 
denominator. 

If  the  number  of  linear  factors  in  numerator  and  de- 
nominator is  the  same,  as  in  ^,  T~=,  the  term  is  said  to 

b  Sea* 

be  of  the  zero  degree.    Such  terms  are  equivalent  to  abstract 
numbers  or  coefficients. 

Terms  of  the  first  degree  always  represent  lengths ;  for 
they  are  always  reducible  to  the  product  of  a  coefficient 
and  a  linear  factor.  (Ex.  4.) 

.    T-,  ab      a  „  ,      3a?b2c       3a?b*  ± 

Examples:  60,   —  =  -Xb,    7^-7  =  7^37 X  c. 


THE   ALGEBRAIC   METHOD.  157 

Terms  of  the  second  degree  always  represent  areas  ;  for 
they  are  always  reducible  to  the  product  of  a  coefficient 
and  two  linear  factors.  (Ex.  7.) 


,     6a3      6a        2     Sa4b3  „  ,2 

Examples  :  nab,    —  =  —  X  a2,    —  —  =  —  —  X  &  . 
b         b  *2         *2 


Terms  of  the  third  degree  always  represent  volumes,  and 
do  not  occur  in  Plane  Geometry. 

Terms  of  degrees  higher  than  the  third  have  no  geomet- 
rical signification. 

The  degree  of  an  irrational  term  (i.e.  a  term  under  a 
radical  sign),  is  found  by  dividing  the  degree,  estimated 
without  regard  to  the  radical  sign,  by  the  index  of  the 
root.  Irrational  terms,  like  rational  terms,  if  of  the  first 
degree,  represent  lengths  ;  if  of  the  second  degree,  areas  ; 
and  if  of  the  third  degree,  volumes. 


Examples  :     A/— r~  is  of  the  first  degree ; 

^1    a      is  also  of  the  first  degree ; 
^   c  a 

is  of  the  second  degree. 


An  algebraic  expression  consisting  of  two  or.  more  terms 
is  said  to  be  homogenous  if  all  the  terms  are  of  the  same 
degree. 

Examples : 

mob      no?         -I  f  ,1      n 

5  a  H —  — j-  is  a  homogeneous  expression  ot  the  first 

degree ; 


V a2  +  b*  -f  c2  +  d2  is  also  of  the  first  degree  ; 
3azb  -  4a&2  -f  653  is  of  the  third  degree. 

The  algebraic  expressions  of  the  values  of  geometric  mag- 
nitudes are  always  homogeneous;  for  all  the  parts  which 


158  GEOMETRY. 


compose  the  whole  must  be  of  the  same  kind  as  the  whole 
(either  lines  or  surfaces  or  volumes),  and  therefore,  the 
terms  which  represent  the  values  of  the  parts  must  be  of 
the  same  degree. 

Homogeneous  expressions  of  the  first  degree  represent 
lengths  ;  of  the  second  degree,  areas ;  of  the  third  degree, 
volumes.  Homogeneous  expressions  of  degrees  higher  than 
the  third  have  no  geometrical  meaning. 

From  the  degree  of  an  expression,  therefore,  it  can  be 
inferred  whether  it  is  capable  of  geometric  construction, 
and,  if  so,  what  kind  of  geometric  magnitude  it  represents. 

The  principle  of  homogeneity  also  affords  one  test  of  the 
correctness  of  expressions.  If,  for  example,  we  obtain  an 
expression  for  a  certain  line  which  is  either  not  homoge- 
neous or  not  of  the  first  degree,  we  may  conclude  at  once 
that  our  investigation  is  somewhere  at  fault. 

§  27.   EXAMPLES  OF  THE  ALGEBRAIC  METHOD. 

I.  To  divide  a  given  line  a  into  two  parts  such  that  the 
difference  of  their  squares  shall  be  equal  to  their  product. 


Analysis  (Fig.  101).  Let  AB  =  a,  and  let  0  be  the  required  point 
of  division;  then  the  problem  is  solved  when  AC  has  been  deter- 
mined. Let  A0=  x\  then  B 0=  a  —  x,  and 

x*-(a-x)*  =  x(a-x); 
whence,  x2  +  ax  —  a2  =  0, 

an  equation  of  the  form  considered  in  Ex.  49.     Its  roots  are 


THE    ALGEBRAIC    METHOD.  159 


Construction.  We  may  proceed  as  in  Ex.  48,  or  as  follows  :  Draw 
BD  _L  AB  and  equal  to  \a,.  Join  AD.  With  D  as  centre  and  DB  as 
radius,  describe  a  circle  cutting  AD  in  E  and  AD  produced  in  F. 
Then,  in  absolute  value,  xl  =  AE,  xz  =  AF.  In  sign  xl  is  positive, 
xz  negative  ;  therefore,  upon  AB  take  A0=  AE,  and  upon  BA  pro- 
duced to  the  left,  take  AG  =  AF. 

Discussion.  The  point  0  falls  between  A  and  B  (Why  ?)  and  is 
evidently  the  only  solution  of  the  problem  in  the  strict  sense  of  the 
enunciation.  If,  however,  the  problem  had  been  stated  as  follows  : 
To  produce  a  given  line  BA  to  a  point  G  so  that  GB2  —  GA2  =  GB 
X  GA  ;  then,  putting  AG  =  x,  the  equation  would  have  been 

(a  +  re)2  —  x2  =  x  (a  +  x), 
the  positive  root  of  which, 


x 


is  identical  in  absolute  value  with  the  negative  root  #2  of  the  equa- 
tion of  the  original  problem.  It  appears,  then,  that  the  algebraic 
solution  of  the  given  problem  supplies  also  the  solution  of  a  closely- 
related  problem,  when  both  roots  of  the  equation  are  considered. 
And  this  suggests  a  mode  of  stating  the  given  problem  in  a  general- 
ized form,  so  as  to  include  both  problems : 

Upon  a  line  passing  through  two  given  points  A,  Bt  to  find  a  point 
G  such  that  the  difference  between  the  squares  of  its  distances  from  A 
and  B  shall  be  equal  to  their  product. 

It  may  be  noticed  that  the  construction  for  the  point  G  (Fig.  101) 
is  the  same  as  that  required  in  the  problem,  To  divide  a  given  line  in 
extreme  and  mean  ratio;  in  fact,  the  proportion  to  be  satisfied  in 
extreme  and  mean  ratio,  a  :  x  =  x  :  a  —  x  gives  the  equation 

a;2  -f  ax  —  a2  =  0. 

II.  In  a  given  square,  to  inscribe  an  equilateral  triangle, 
having  one  vertex  in  common  with  the  square,  and  the 
other  two  vertices  in  the  sides  of  the  square. 


160 


GEOMETRY. 


Analysis.  (Fig.  102.)  Let  ABCD  be. the  given  square,  AEF  the 
triangle  required.  If  AB  =  a,  BE  =  x,  then  AE2  =  a2  +  x2,  and 
lEF2  =  2(a-  x)2.  But  AE  =  EF;  therefore,  a?_+  a;2  =  2 (a  -  a;)2. 
Solving  this  equation  for  x,  we  have  x  =  2  a  ±  a  V3. 

Both  values  of  x  are  positive,  but  the  greater  value  is  greater  than 
a,  and  must  be  rejected  if  the  sides  of  the  square  are  regarded  as  lines 
limited  by  the  vertices. 

A  D 


\ 


\ 


\ 


v 


8 


C  G 

Fig.  102. 

Construction.  Produce  BOio  G,  making  CG  =  a.  Join 
DB,  upon  DB  take  DH=DA,  and  join  GH;  then  GH=aV3. 
Upon  GJB  take  GE  =  #J7 ;  then  £#=  2  a  -  a  V3,  and  Eis  a  vertex 
of  the  triangle  required.  From  A  as  a  centre,  with  AE  as  radius, 
cut  CD  in  J7!  Join  EF.  AEF  is  the  triangle  required. 

A  proof  that  AEF  is  an  equilateral  triangle  may  readily  be  given, 
and  is  left  as  an  exercise  for  the  learner. 

Discussion.  The  construction  of  the  other  root  2a  +  a  V3~  deter- 
mines a  point  K  in  BC  produced.  If  L  denote  a  point  in  DC  pro- 
duced such  that  AL  =  AK,  it  is  easy  to  prove  that  the  A  AKL  will 
also  be  equilateral.  The  A  AKL,  therefore,  is  also  a  solution  of  the 
problem,  provided  the  sides  of  the  square  are  regarded  as  lines  un- 
limited in  length. 

III.  Having  given  a  circle  and  a  point  P,  to  draw 
through  P  a  secant  such  that  the  exterior  segment  shall  be 
equal  to  the  interior  segment. 

Analysis.  Let  0  be  the  centre  of  the  given  circle,  r  its  radius, 
OP  =  a;  then,  if  x  denote  the  exterior  segment,  we  have  (No.  163), 


whence, 


THE   ALGEBRAIC    METHOD.  161 

The  Construction  and  Proof  present  no  difficulties  requiring  ex- 
planation. 

Discussion.  The  negative  value  of  a;  corresponds  to  no  closely 
related  problem,  and  must  be  rejected.  The  problem  from  its  nature 
is  impossible  either  if  a  <  r,  or,  if  x  >  2  r ;  that  is,  if  a  >  3  r ;  in  the 
first  case,  the  impossibility  is  indicated  by  the  value  of  x  becoming 
imaginary. 

IV.  From  two  given  lines  a,  b,  to  take  away  equal  parts 
in  such  a  way  that  the  sum  of  the  remainders  shall  be  equal 
to  a  given  line  c. 

Analysis.  Let  x  denote  one  of  the  equal  parts ;  then,  by  the 
conditions, 


whence,  a;  = 

The  Construction  is  obvious. 

Discussion.  If  c  is  greater  than  a  +  6,  the  value  of  x  is  negative. 
In  order  to  see  how  the  statement  of  the  problem  should  be  modified 
to  cover  this  case,  substitute  —  x  for  x  in  the  original  equation ;  it 
now  takes  the  form  (a  +  x)  +  (b  +  x)  =  c,  which  is  the  direct  algebraic 
expression  of  the  problem :  to  add  equal  parts  to  two  given  lines  so 
that  their  sum  shall  be  equal  to  a  third  line. 

C"  A  C  B  C' 


Fig.  103. 

V.  The  intensities  at  unit  distances  of  two  luminous 
points  A,  B  are  represented  by  a,  5,  respectively ;  to  find 
a  point  C  in  the  line  AB,  which  is  equally  illuminated  by 
the  two  points  of  light. 

Analysis.  (Fig.  103.)  Let  AB  =  d,  AC  =  x;  then,  since  by  a  law 
of  optics  the  quantities  of  light  received  at  C  from  A  and  B  are  in- 
versely as  the  squares  of  the  distances,  we  have, 


162  GEOMETRY. 


This  equation,  if  we  put  -  =  w2,  becomes 

(d  —  .r)2  =  m?x2  ; 
whence,  d  -  x  =  +  mx. 

If  ojj,  x2  denote  the  true  values  of  x,  we  have, 

d 


d 


and  it  appears  that  in  general  there  are  two  points  equally  illuminated 
in  the  line  AB. 

The  Construction  presents  no  difficulty. 

Discussion.  The  expression  for  x2  gives  rise  to  three  cases,  accord- 
ing as  m  is  less  than,  equal  to,  or  greater  than,  1. 

CASE  I.  m<l.  Then  x-^  and  x%  are  both  positive,  xl  is<e?,  but 
#2  is  >  d.  Therefore,  there  is  one  point  0  between  A  and  B,  and 
another  point  C1  to  the  right  of  B.  This  is  otherwise  evident;  for, 
if  m  is  <  1,  then  b  is  <  a,  that  is,  B  is  the  feebler  light,  and  the  two 
points  of  equal  illumination  must  both  be  nearer  to  B.  In  this  case, 
BC=  x  —  d,  but  the  squares  of  x  —  d  and  d  —  x  being  equal,  equation 
(1)  still  holds  true. 

CASE  II.   ra=l.   Then  xl  =  -,  x%  =  oo.    This  result  means  that  the 

point  Cis  now  halfway  between  A  and  B,  and  that  the  point  C"  no 
longer  exists. 

CASE  III.  m  >  1.  In  this  case  xl  is  positive  and  less  than  d,  while 
x2  is  negative.  The  points  equally  illuminated,  therefore,  are  a  point 
C  between  A  and  B,  and  a  point  C"  to  the  left  of  A.  This  result  is 
otherwise  an  evident  consequence  of  the  fact  that  now  A  is  the  feebler 
light.  If  in  equation  (1)  we  write  —x  for  re,  the  equation  becomes 

5-      b 

x*     (d  +  x? 

But  this  form  corresponds  exactly  to  the  position  of  C",  since, 
in  this  case,  C"  B  =  d  +  x. 

Thus,  equation  (1)  covers  all  possible  cases  in  the  solution  of  the 
problem,  provided  that  we  regard  x  as  positive  or  negative,  according 
as  the  point  C  is  situated  to  the  right  or  to  the  left  of  A. 


THE   ALGEBRAIC    METHOD.  163 

The  foregoing  examples  show  how  Algebra  may  be  applied 
in  the  solution  of  problems,  and  what  the  advantages  of  the 
algebraic  method  are.  By  an  algebraic  analysis  we  obtain 
for  the  value  of  an  unknown  length  a  general  expression 
which  includes  all  possible  hypotheses  respecting  the  val- 
ues of  the  given  lengths;  so  that  a  mere  examination 
of  this  expression,  without  a  figure,  usually  suffices  to  deter- 
mine the  limits  of  possibility  of  the  problem.  By  introduc- 
ing the  conception  of  negative  lines  the  algebraic  method 
enables  us  to  group  together  related  problems,  to  see  clearly 
their  exact  relations,  and  to  bring  them,  when  possible,  un- 
der a  single  generalized  form  of  statement. 

In  applying  this  method,  the  following  principles  must 
be  observed : 

I.  As  many  equations  are  necessary  as  there  are  unknown 
lengths  (usually  only  one)  to  be  determined. 

II.  An    imaginary    value,    obtained    for   an   unknown 
length  x  indicates  that  the  problem  is  impossible. 

III.  A  negative  value  of  x  must  be  rejected  if  x  denote 
simply  an  absolute  length  without  reference  to  position  or 
direction  (e.g.  the  radius  of  a  circle). 

IV.  A  negative  value  of  x,  when  the  position  or  direc- 
tion of  x  enters  into  account,  usually  corresponds   to   a 
modified  or  extended  statement  of  the  problem  ;    to  see 
what  this  should  be,  substitute  — x  for  x  in  the  original 
equation  and  then  translate  the  new  equation    into  the 
statement  of  a  problem. 

V.  If  a  positive  value  of  a  length  x  is  properly  repre- 
sented by  a  line  drawn  from  a  given  point  in  a  certain 
direction,  then  a  negative  value  of  x,  if  admissible  from  the 
nature  of  the  question,  will  be  represented  by  a  line  drawn 
from  the  same  point  in  the  opposite  direction.     (Principle 
of  Descartes.) 


164  GEOMETRY. 


§  28.    PROBLEMS. 
GROUP  I.    Classified  Equations  of  the  First  Degree. 

A.   Expressions  of  the  form  x  =  a  ±  b  ± ,  or  x  = — - — '• 

n 

1.  To  divide  a  triangle  into  two  parts   having  equal 
perimeters  by  drawing  a  line  from  one  vertex. 

2.  To  divide  a  parallelogram  by  a  line  drawn  from  one 
vertex  into  two  parts  such  that  the  perimeter  of  one  shall 
exceed  that  of  the  other  by  a  given  length  a. 

3.  To  take  equal  lengths  from  two  sides  of  a  given  tri- 
angle, measured  from  their  intersection,  so  that  the  sum  of 
the  remainder*  shall  be  equal  to  the  third  side. 

4.  To  describe  a  circle  about  each  of  three  given  points 
so  that  each  circle  shall  touch  the  other  two  externally. 

5.  To  draw  parallels  at  equal  distances,  respectively, 
from  the  sides  of  a  given  rectangle,  so  that  they  shall  form 
another  rectangle  of  given  perimeter. 

6.  To  divide  the  greater  side  of  a  given  rectangle  into 
two  segments,  so  that  the  difference  of  their  squares  shall 
be  equal  to  the  area  of  the  rectangle. 

7.  To    divide    a   given   trapezoid   into   two    equivalent 
quadrilaterals  by  drawing  a  line  parallel  to  one  of  the  legs. 

8.  To  divide  one  side  of  a  given  triangle  into  two  parts 
such  that  their  difference  shall  be  equal  to  one-third  of  the 
sum  of  the  other  two  sides. 

B.   Expressions  of  the  form  x  =  — • 

9.  Upon  one  side  of  a  given  triangle,  to  construct  a  rec- 
tangle equal  in  area  to  the  product  of  the  other  two  sides. 


THE   ALGEBRAIC    METHOD.  165 

10,  Given  a  point  P  in  the  side  AB  of  the  rectangle 
ABCD;  through  Pto  draw  a  line  meeting  the  side  BC 
produced  in  a  point  X,  so  that  the  triangle  PBX  shall  be 
equivalent  to  the  rectangle. 

Given  a  triangle  ABC;  to  draw  a  line  XY  parallel  to 

AB  (X'm  AC,-  Fin  BC)  so  that  one  of  the  following  con- 
ditions shall  be  satisfied  : 

11,  XY=BY.  17,   AX+BY=d. 

12,  CX  =  BY.  18.   AX-BY-^d. 

13,  XY=AX+BY.  19.   CX+BY=d. 

14,  XY=BY-AX.  20,   BY-OX  =  d. 

15,  XF=  <7X  -{-  .5  F.  21,   ^+JTF=<7X+<?F 

16,  XY=CX-BY.  22,   4J3-  XF=  CX+  CF. 


23,  Through  a  given  point  in  one  side  of  a  triangle,  to 
draw  a  line  cutting  off  a  triangle  equal  in  area  to  one-third 
of  the  given  triangle. 

24,  Upon   a   given   line   are   four  points  in  the  order 
A,   C,   D,  B]  to  find  in  the  line  a   point  X  such  that 
AX:BX=DX:  CX. 

25,  Upon  a  given  line  as  one  side,  to  construct  a  rec- 
tangle equivalent  to  a  rhombus  whose  diagonals  are  equal, 
respectively,  to  two  given  lines. 

C.    Expressions  of  the  form  x  =  —  • 

b 

26,  Upon  a  given  side,  to  construct  a  rectangle  equiva- 
lent to  a  given  square. 

Given  a  triangle  ABC]  to  draw  a  line  XY  parallel  to 
AB,  so  that  one  of  the  following  conditions  are  satisfied  : 

27,  XF2  =  CXx  AC.     (Let  x  =  XY.*) 


166  GEOMETRY. 


28,  XY  =?=  CX-  AX.    (Let  x  =  CX.) 

29,  XY  =  AX-  CX.    (Let  x  =  CX.) 

30,  To  produce  a  given  chord  of  a  circle,  so  that  the 
tangent  drawn  from  its  extremity  shall  have  a  given  length. 

31.  To  transform  a  given  square  into  a  triangle,  having 
a  given  altitude  and  a  given  angle  at  the  base. 

32,  To  produce  a  given  line  AB  to  a  point  X,  such  that 
the  product  AB  X  AX  shall  be  equal  to  a  given  square. 

D.   Expressions  of  the  form  x  =  ^— = — 

no 

33.  To  construct  a  rectangle,  given  one  side  and  the  sum 
of  the  diagonal  and  the  other  side. 

34,  To  construct  a  rectangle,  given  one  side  and  the  dif- 
ference of  the  diagonal  and  the  other  side. 

35.  Upon  one  side  of  a  given  triangle,  to  construct  a 
rectangle  equivalent  to  the  sum  of  the  squares  of  the  other 
two  sides. 

36,  Given  a  point  P  in  a  diameter  of  a  given  circle 
(centre   0) ;  to  produce  the  diameter  to  a  point  X,  such 
that  the  tangent  drawn  from  X  to  the  circle  shall  be  equal 
to  PX.     (Let  a=OP,x  =  OX.) 

GROUP  II.     Unclassified  Equations  of  the  First  Degree. 

To  divide  a  given  line  into  two  segments  such  that  : 

37.  The  difference  of  their  squares  shall  be  equal  to  52. 

38.  Their  ratio  shall  be  equal  to  that  of  two  given  squares. 

39.  Their  ratio  shall  be  equal  to  that  of  two  given  rec- 
tangles. 


THE    ALGEBRAIC    METHOD.  167 

Given  a  point  C  in  a  line  AB  ;  to  find  iii  this  line  a 
point  X  such  that  : 

40, 


41,    AX\  BX  =  AC2  :  BC\ 

Given  a  triangle  ABC;  to  draw  a  line  XY  parallel  to 
BC,  so  that  one  of  the  following  conditions  shall  be  satis- 
fied : 


42,  XY*  =  ACx  CY. 

43,  XYZ=  OYX  BY. 

44,  XY'2  =  AXx  CY. 

45,  AX2  -  A  F2  =  XYx  BO. 

46,  So  that  the  two  parts  of  the  triangle  ABC  shall 
have  equal  perimeters. 

47,  So  that  the  perimeter  of  the  triangle  ^4X]Fshall  be 
equal  to  half  that  of  the  trapezoid  BCYX. 

To  inscribe  in  a  given  triangle  : 

48,  A  square. 

49,  A  rhombus,  given  one  of  its  angles. 

50,  A  rectangle,  given  the  ratio  of  its  sides. 

51,  A  rectangle,  given  its  perimeter. 

52,  A  rectangle  with  one  diagonal  parallel  to  one  side 
of  the  triangle. 

53,  A  parallelogram  of  given  perimeter,  and  having  one 
angle  in  common  with  the  triangle. 

54,  To  divide  a  trapezoid  by  a  line  parallel  to  the  bases 
into  two  parts  having  equal  perimeters. 


168  GEOMETRY. 


55,  To  circumscribe  a  trapezoid  about  a  given  circle, 
given  the  lengths  of  the  two  legs.     Find  also  the  area  of 
the  trapezoid  in  terms  of  the  given  parts. 

To  construct  a  right  triangle,  given  : 

56,  a,  p.  59,    a  +  b  -f  c,  A.  61,    c  +  a,  b. 

57,  c  —  a,  p.          60,    a  +  b  —  c,  h.  62,    c  —  a,b. 

58,  a  -f  5,  p.     (In  59-62  compute  the  hypotentlse  c.) 

In  one  side  of  a  given  triangle,  to  find  a  point  Such  that 
the  perpendiculars  dropped  from  this  point  to  the  other 
two  sides  : 

63,  Shall  have  a  given  sum. 

64,  Shall  be  in  a  given  ratio. 

65,  In  a  given  square  to  inscribe  five  equal  circles,  so 
that  the  middle  circle  shall  touch  the  four  others,  and  each 
of  these  four  shall  touch  two  adjacent  sides  of  the  square. 

66,  Given  P  in  L,  and  Q  not  in  L  ;  to  find  a  point  X 
in  L,  so  that  PX+  QX=  a. 


67,  Given  P  in  L,  and  a  second  line  L1  ;  to  find  a  point 
X  in  L,  so  that  the  sum  of  its  distances  from  P  and  from 
If  shall  be  equal  to  a  given  length  a. 

68,  Given  a  rectangle  ;  to  construct  a  square  such  that 
the  perimeters  of  the  two  figures  shall  have  the  same  ratio 
as  their  areas. 

69,  To   draw   a   line   parallel   to   one   side  of  a  given 
triangle  which  shall  cut  off  a  trapezoid  having  a  given 
perimeter. 


THE   ALGEBRAIC    METHOD.  169 

GROUP  III.    Pure  Quadratic  Equations. 

70,  Giyen  O  in  AB\  to  find  X  in  AB,  so  that  AX* 
=  ABxAC. 

71,  In  a  triangle  ABC,  to  draw  a  line  XY between  AC 
and  BQ{  so  that  the  triangle  GXY  shall  be  isosceles  and 
equal  in  area  to  half  the  triangle  ABO. 

72,  To  transform  a  given  parallelogram  into  a  rhombus, 
one  angle  remaining  unchanged. 

73,  To  construct  a  circle  which  shall  pass  through  two 
given  points,  and  shall  touch  a  given  line. 

To  divide  a  given  triangle  by  a  line  parallel  to  one  side 
into  : 

74,  Two  equivalent  parts. 

75,  Two  parts  in  the  ratio  2  :  3. 

76,  Within  a  given  circle  to  construct  five  equal  squares, 
so  that  each  of  the  four  outer  squares  shall  have  two  ver- 
tices in  the  circumference,  and  one  side  in  common  with 
the  fifth  square. 

77,  To  construct  a  right  triangle,  given  c,  and  a  =  2b. 

78,  To  construct  a  square  which  shall  be  to  a  given 
square  as  3  :  2. 

To  construct  two  lines  (x  and  y)  given  : 

79,  Their  ratio  and  their  product. 

80,  Their  ratio  and  the  difference  of  their  squares. 

81,  Their  sum  and  the  difference  of  their  squares. 

82,  To  transform  a  given  square  into  a  rectangle  whose 
sides  are  as  3  :  4. 


170  GEOMETRY. 


Given  a  length  a,  a  circle,  and  an  exterior  point  P  ;  to 
draw  a  secant  PAB  cutting  the  circumference  in  A,  B,  so 
that: 

83,   PAxAB  =  a\  84,   PBxAB  =  a\ 

85,    Given  a  length  a,  a  circle,  and  an  interior  point  P; 
to  draw  through  P  a  chord  meeting  the  circumference 
A,  B,  so  that  AB  X  PB  =  a\ 


n 


To  divide  a  given  line  AB  in  JTso  that  : 
86,   ABz-AX2  =  b\          87, 


To  divide  a  given  trapezoid,  by  a  line  parallel  to  the 
bases,  into  : 

88,  Two  equivalent  parts. 

89,  Two  parts  in  a  given  ratio  m  :  n. 

90,  To  cut  from  a  given  triangle,  by  a  line  parallel  to 
one  side,  a  trapezoid  of  given  area. 

GROUP  IV.    Complete  Quadratic  Equations  Classified. 
A.    Equations  of  the  form  a;2  ±  ax  —  a?  =  0. 

91,  To  divide  a  given  line  in  extreme  and  mean  ratio. 

92,  To  divide  a  given  line  into  two  parts,  such  that 
the  smaller  shall   be   a   mean   proportional   between   the 
greater  and  the  difference  of  the  two. 

93,  Given  a  circle  and  a  chord  ;  to  produce  the  chord  to 
a  point  X,  so  that  the  tangent  drawn  from  X  shall  be  equal 
to  the  chord. 

94,  In  the  triangle  ABO  to  draw  XY  parallel  to  AB, 
so  that  AB  X  AX=  XYx  CX. 


THE   ALGEBRAIC    METHOD.  171 

95,  Upon  a  given  line  as  hypotenuse  to  construct  a  right 
triangle  in  which  the  greater  leg  shall  be  a  mean  propor- 
tional between  the  hypotenuse  and  the  other  leg. 

To  produce  a  line  AB  to  a  point  X,  so  that  : 

96,  ABz  =  AXxBX.        97, 


B.  Equations  of  the  form  x2  ±  ax  —  62  =  0. 

98.    To  transform  a  given  square  into  a  rectangle,  having 
given  the  difference  of  its  sides. 

Given  the  point  G  in  the  line  AB\  to  find  the  point  X 
in  AB,  so  that  : 

99,   ABxBX=AXxCX. 

100,  AX*  +  BX*-CX*  =  AC\ 

101,  Given  a  circle  and  a  chord  ;  to  produce  the  chord 
to  a  point  JT,  so  that  the  tangent  drawn  from  X  shall  have 
a  given  length. 

102,  To  construct  a  circle  which  shall  touch  a  given  line 
and  cut  a  chord  of  given  length  from  another  given  line. 

103,  Through  a  given  point  within  a  given  circle  to  draw 
a  chord  so  that  the  difference  of  the  two  segments  shall  be 
be  equal  to  a  given  line. 

104,  To  divide  a  given  line  into  two  parts,  so  that  the 
square  of  one  part  shall  be  equal  to  half  the  square  of  the 
other  part. 

105,  To  construct  a  right  triangle,  given  a  —  b,  h. 

C.  Equations  of  the  form  x2  ±  ax  —  be  —  0. 

106,  To  divide  a  given  line  into  two  parts,  so  that  the 
square  of  one  part  shall   be  equal  to  the  product  of  the 
other  part  and  a  given  length. 


172  GEOMETRY. 


Given  C  in  AB  ;  to  find  X  in  AB,  so  that  : 
107,    OX*  =  A  C  X  J3  JT.        108,    v4X'  =  AB  x 


109,  To  transform  a  given  square  into  an  isosceles  tri- 
angle in  which  the  difference  of  the  base  and  altitude  is 
given. 

110,  A  semi-circumference  is  divided  by  a  line  perpendic- 
ular to  the  diameter  into  two  unequal  parts  ;   to  construct 
a  circle  which  shall  touch  the  diameter,  the  perpendicular, 
and  the  smaller  part  of  the  semi-circumference. 

D.    Equations  of  the  form  #2  ±  ax  +  62  =  0. 

NOTE.  If  b  >  |  a,  the  solution  is  impossible.  If  b  =  Ja,  x  =  ±  \a. 
Therefore,  b  =  $  a  is  the  maximum  value  of  b,  if  b  be  regarded  as  a 
variable  quantity;  and  a  =26  is  the  minimum  value  of  a,  if  a  be 
regarded  as  a  variable  quantity. 

111,  To  divide  a  given  line  into  two  parts  the  product 
of  which  shall  be  equal  to  the  square  of  a  given  line. 
When  is  this  product  a  maximum  ? 

112,  To  construct  a  rectangle,  given  its  perimeter  and  its 
area.     Of  all  rectangles  of  given  perimeter  which  has  the 
greatest  area?     Of  all  rectangles  of  given  area  which  has 
the  least  perimeter? 

113,  To  inscribe  in  a  given  equilateral  triangle  another 
equilateral  triangle  one-half  as  large. 

114,  Given  C  in  AB  ;  to  find  a  point  X  in  AB,  such 
that  GX1  =  AXx  BX. 

115,  In  a  given  isosceles  right  triangle  to  inscribe  a  rec- 
tangle having  a  right  angle  in  common  with  the  triangle, 
and  equivalent  to  a  given  square. 


THE   ALGEBRAIC    METHOD.  173 

E.    Equations  of  the  form  a2  ±  ax  +  be  =  0. 
NOTE.    As  regards  maxima  and  minima,  see  note  under  D. 

116,  To  divide  a  given  line  into  two  parts,  so  that  their 
product  shall  be  equal  to  the  area  of  a  given  rectangle. 

117,  Given  C  in  AB  ;  to  find  a  point  X  in  AB  between 
B  and  <7,  so  that  A X*  =  BXx  CX. 

118,  Given  0  in  A  B ;  to  find  a  point  X  in  AB  between 
A  and  (7,  so  that  AB  :  AX=  BX:  CX. 

119,  To  construct  a  rectangle  equivalent  to  a  given  rec- 
tangle, and  having  a  perimeter  equal  to  that  of  another 
given  rectangle. 

120,  To  construct  a  right  triangle,  given  a  +  b,  c. 

121,  To  construct  a  right  triangle,  given  a~b,  c. 

122,  To  divide  the  greater  side  of  a  given  rectangle  into 
two  parts,  so  that  the  sum  of  their  squares  shall  be  equal  to 
the  area  of  the  rectangle. 

123,  To  transform  a  given  square  into  an  isosceles  tri- 
angle in  which  the  sum  of  the  base  and  the  altitude  is 
given. 

124,  In  a  given  square  to  inscribe  another  square  the 
side  of  which  has  a  given  length. 

125,  To  transform  a  given  triangle  into  a  rectangle  hav- 
ing the  same  perimeter. 


ANSWERS 

TO    TILE 

NUMERICAL   EXERCISES  IN   CHAPTERS   III.,   IV.,  V. 


CHAPTER 

III.     §  14. 

1.  16.666ft. 

12.  EF=  2287  ft. 

2.  54  ft. 

13     ^  . 

3.  35.49  ft. 

4.  122  feet,  183  feet. 

14,  90  ft. 

5.  Yes. 

15,  320  ft. 

6.    4  and  5,  4.5  and  7.5, 

16.  4^-  miles. 

6.43  and  8.57. 

17.  8.57  ft. 

„      ac          be     .  c 

18.  26  ft. 

a  +  b'    a  +  b  '  2 

19.  20  ft. 

8.  2:1. 

CL 

9.  60,  56.25. 

20.  7;V3=:0.866xa. 

10,  AB  =  8  ft., 

21,  ^  =  4.8  ft.,  y  —  6.4ft. 

AD=  10.66  ft., 

22,  x  =  3.846  ft., 

DE=  5.33  ft. 

y-  22.  154  ft., 

11.  16.25  in.  ;  8.75  in. 

h  -9.231  ft. 

OQ     ~                ft2                                   ^2 
40.   ju                         >     y 

i           ab 
—  >     fi  • 

24.  a:  =  0.6,  y  =  0.8. 

25.  3,  4,  5.     The  second  solution  (—1,  0,  +1)  does  not 

belong  to  this  question. 

26.  24,  32.  28.  12  ft. 

27.  33,  44,  55.  29.  2.238,  3.059. 


ANSWERS.  175 


30,  8  in.  and  2  in.  32,   12  in. 

31.  rV3,  120°.  33. 

34,  V200=14.142in. 

35.  Each  tangent  =  8  in. ;  chord  of  contact  =  9-|  in. 

36.  r  =  — - —  ;  r  —  a,  and  ab  is  a  diameter. 

2a 

37.  Obtuse,  since  82  +  92<  132.     (No.  163.) 
38          a*  40-  See  text. 

'  2V^^'  41,  7  in.  and  4|  in. 

39.   Va6  +  c2.  42,  10  in.  and  2  in. 

43.  14.14  ft.  44.  rVIO.  45.  6  in. 


CHAPTER  IV.     §  20. 

1.  5  acres  1400  sq.  yds.  18.  The  rectangle,  8" 

2.  2.25<im.  19.  18  ft.  and  12  ft. 

3.  39  sq.  ft.  9  sq.  in.  20.  30  ft.  and  10  ft. 

4.  648  sq.  yds.  21,  12  ft.  and    5  ft. 
.5.  220  yds.  22.  42  ft.  and  18  ft. 

6.  200  rds.  23.  16  ft.  and  12  ft. 

7,  $60.48.  24.  2700"m. 

8.  $166.50.  25,  60  ft.  and  40  ft. 

9,  7.789  yds.  26,  4  ft. 

10.  49  sq.  ft.  54  sq.  in.  27,  7168  tiles. 

11.  7  acres  3920  sq.  yds.  28.  2  ft.  4  in. 

12.  8748  sq.  yds.  29.  $558.32. 

13.  11  yds.  30.  Gained  $3518. 

14.  97  yds.  2^  ft.  31.  No  difference. 

15.  288  sq.  ft.  32.  11,250  sq.  ft. 

16.  25  sq.  ft.  33.  96qm. 

17.  225  sq.  yds.  34.  108  sq.  ft. 


176  GEOMETRY. 


35,  16J  sq.  ft.  54,  216  sq.  ft. 

36,  20  ft.  55,  1.732  sq.  ft. 

37,  No  ;  1:2.  56,  692.82  sq.  ft. 

38,  1 :  2.  57,  332.544  sq.  ft. 

39,  17  ft.  and  20  ft.  58,  3  :  4. 

40,  33.294m.  59,  200  ft. 

41,  75  yds.,  72  yds.  60,  5240  sq.  ft.  nearly. 

42,  1164  sq.  ft.  54  sq.  in.  61,  44.7213m. 

43,  1 :  3.  62,  150  sq.  ft. 

46,  a :  a1  =  h' :  h. .  63,  150  sq.  ft. 

47,  They  are  equal.  64.  216  sq.  ft.  and  384  sq.  ft. 

48,  448  sq.  miles.  .      65,  70.9298qm. 

49,  30  sq.  ft.  66,  48.441m  and  12.386m. 

50,  162  sq.  ft.  67,  Base  70  ft.,  altitude  25  ft. 

51,  100  sq.  ft.  68.  147-J-  sq.  ft. 

52,  30  sq.  ft.  69,  5460  sq.  ft. 

53,  5.196  sq.  in.  70,  420  sq.  ft.       71,  0.38  ft. 

72,  Lets=£+£-fc      then 


r  ^/O* -«)(*-  *)  (*-<0  ; 
S 

abc 


73.  16.8  ft.  80.  208  sq.  ft. 

74.  7.49qm.  81.  4  sq.  ft. 

75.  1050  sq.  ft.  82.  28  ars  ;  33£  ars. 

76.  300  sq.  ft.  85,  $417.60. 

77.  20  ft.  86.  4896  sq.  ft. 

78.  (i.)  13£  ft.  and  161  ft.    87,  5000  sq.  ft. 

(ii.)  12  ft.  and  18  ft.  88,  6  acres  6440  sq.  ft. 

79.  $1600.  89.  80  ft. 


ANSWERS.  177 


90,  160ft. 

99.  18.2363ft. 

91,  40  ft. 

100.  14.7418ft. 

92,  ISft-  ft- 

101,  8.658  ft. 

93,  40  ft. 

102,  10.528  ft. 

94,  27ift. 
95,  18ft. 
96,  3^  ft. 
97,  12.7279ft. 

103,  26.61  ft, 
104,  62.161  yds. 
105,  80  ft. 
106,  30  ft,  and  45 

ft. 

98,  7.8964  ft. 

107,  12  ft.  and  8  ft. 

108.  (i.)  i  and  f  ; 
(iv.)  -^  and  | 

(vi.)    —  and 

(ii.)  1  and  |  ;    (iii.)  i  and  f  ; 

f      \                      -\    (YYUYL        -L 

4  ;      (v.)  —  .  and  -      —  ; 
mn              mn 

109,  %  and  £  or  f  and  f,  etc.  110,  60  ft. 

111,  One  line  to  the  adjacent  side  dividing  it  in  the  ratio 

1  :  2,  the  other  line  to  the  opposite  vertex. 

112,  AE=:&AC,  BF=^BC. 

113,  AD  =VJ  X  AB.  118,  \. 

114,  AD  = VJ  X  AB.  119,  5  :  1. 

115,  AD  = Vf  X  AB.  120<  T72- 

/-^—  121,  2:1. 

116,  AD=\I—  —X  AB.      19o     A  -p—  Q  f \  q  in 

\^^_i_^2,  ifla,  -d.-T  —  o  it.  u  in. 

117,  ^4^=  21.21  ft.  123,  DX :  XB  =  8: 17. 

124,  9.87  ft. ;  0.564  X  h. 

125,  Upper  part  =  ^-(a  +  56)  ; 

lo 

Middle  part-  A(a-f  &); 

D 

Lower  part   =  —  (5  a  +  b). 
18 


178 


GEOMETRY. 


126. 
127, 
128. 


1250  sq.  ft. 
40  sq.  ft. 
60ft. 


£ 

V2 


129.  -. 

130. 
131. 


V5_ 
2VII. 


140.  Bases,  a  -J >    and  a  — 


132.  21.6333ft. 

133.  5  ft. 

134.  65ft. 

135.  2  :  1. 

136.  (i.)4:l;  (ii.)2:l. 

137.  4:3. 

138.  311  ft.,  52|  ft. 

139.  9ft. 

4 


Each  leg  =  a. 


141. 
142. 

143. 
144. 
150. 
152. 

153. 
154. 

158. 
159. 


F= 


2  Va2  - 


145. 
146. 
147. 
148. 

149. 


a  =2* 


151. 

(i.)  ^(hm  +  an  +  mn).       (iii.)  ^(hm  —  an  —  raw) 
(ii.)  J-  (Jim  +  an  —  m)n.       (iv.)  ^  (an  —  km  —  mri) 

(i.)  F=  (3  -  2V2)  s\      155. /  = 
(ii.)  ^=  (3  +  2V2)  a72.     156 
mw,  (a  —  n)  m, 

/7  \  f  \/7,  \1  Cf? 

(u — VYi)n    (a — n)(o — on).    JLO/ . 
|(V4a2  +  2^±  V4^ 


l 

\ 


+  W+J9 


ANSWERS. 


179 


CHAPTER 

V.     §25. 

22.  c  =  2a,     c  = 

=  2r,    c  =  ^. 

3 

23.  6^2-  V3  -=3.1048  sq.  ft. 

24.   12r2(2-V3). 

47.  4  ft.  4|  in. 

25.  8:3. 

48.  16Jft. 

26.  f 

49.  7ft.  4  in. 

27.  70.53ft. 

50.  30.86m. 

28.  2r(V2-l). 

51.  6369.42k. 

52.  1:2. 

-V/T  A      1^   O-/K 

C  *  -LU  ~p  Zt  v  O 

53.  4  ft. 

29.             4 

30.  4.902  ft. 

54.  1  sq.  ft.  10  sq.  in. 

31.  13  ft. 

55.  203  acres  3080  sq.  yds. 

56.  192.249yds. 

32.  10  acres  1843  sq.  yds., 
nearly. 
33.  $216. 

J 

57.  352yds. 
58.  308  sq.  ft. 

59.  49.638  ft. 

34.  Octagon,  by  754.4  sq.  ft. 

60.  154  sq.ft. 

36.  1  ft.  10  in. 

61.  $107.86. 

37.  4  ft.  1  in. 

62.  2.984yds. 

38.  120  revolutions. 

63.  39.597  ft. 

39.  3  ft.  9^  in. 

64.   1050  sq.  in. 

40.  17^  ft. 

65.  28.284  in. 

41.  220ft. 

66.  5544  sq.  ft. 

42.  11  in. 

67.  240  sq.  ft.  90  sq.  in. 

43.  45°. 

68.  Square,  1089  sq.  ft.  ; 

44.  57^-°. 

Circle,    1386  sq.  ft.  ; 

45^  57T3_°. 

Triangle,  838.312  sq.  ft. 

46.   14  ft.  U  in. 

69.  15ft. 

180 


GEOMETRY. 


70.  Vr2-r'2. 

71.  47r:3V3. 

72.  1:4. 

73.  Circle,  2.122qm; 
Triangle,  0.877qm. 

74.  $376.24f 

75.  5.196  ft,  7.348  ft. 

76.  $2461.85. 

77.  41fft. 

78.  4ft. 

79.  1  ft.  3  in. 

80.  2  ft.  1  in. 

81.  102ft. 

82.  240ft. 

83.  14.277ft. 

84.  78^  sq.  ft. 

85.  33.312  in. 

86.  107.274  ft,  225.868  ft. 

87.  222.562ft. 

88.  616  sq.  ft. 

89.  35  ft,  92  ft. 

90.  578.6625  sq.  ft. 

91.  546.175  sq.  ft. 

92.  442^  sq.  ft. 

93.  342^  sq.  ft. 

94.  2066|  sq.  ft. 

113.  r2 


95.   - 


-3V3) 


100. 


r2(47r-5VlO-2V5) 
40 

101.  STT  sq.  ft. 

102.  12.092  sq.  ft. 
107T  +  3V3 


103. 


12 


104.  4*~  3V^  =  Q. 

87T+3V3 

105.  4.43qm  and  149.5qm. 

106.  27°  7' 18". 

107.  28.2  ft.  and  45.13  ft. 

108.  7ift. 

109.  2ft. 

110.  2.414ft. 

111.  ?(V  +  &2). 


112.  £ 


2  —  '   (^ 

115.  (i.)  7rr2(3  -  2  V2) ; 


116 


EXERCISES   IN   SOLID   GEOMETRY. 


NOTE.     When  TT  occurs  as  a  factor,  take  TC  =  2?2-,  if  English  units 
are  used,  and  TT  =  3.1416  if  Metric  units  are  used. 

The  weight  of  1  cubic  foot  of  water  is  to  be  taken  as  1000  oz. 


§  1.   PLANES. 

1,  How  many  perpendiculars  in  space  can  be  erected  at 
a  given  point  in  a  given  line  ?     How  do  they  lie  relatively 
to  one  another  ? 

2,  How  many  perpendiculars  in  space  can  be  dropped 
from  a  given  point  to  a  given  line  ? 

3,  If  two  lines  in  space  are  both  perpendicular  to  a  given 
third  line,  are  they  necessarily  parallel  to  each  other? 

4,  Can  two  lines  not  in  the  same  plane  both  be  perpen- 
dicular to  a  given  plane  ? 

5,  If  two  lines  in  space  are  both  perpendicular  to  a  given 
plane,  are  they  necessarily  parallel  to  each  other  ? 

6,  If  two  lines  in  space  are  parallel  to  a  given  plane,  are 
they  necessarily  parallel  to  each  other  ? 

7,  Can   two  vertical  planes   intersect?   two   horizontal 
planes  ? 

8,  Through  a  vertical  line  how  many  vertical  planes  can 
be  passed  ?  how  many  horizontal  planes  ? 


EXERCISE    MANUAL. 


9,  Through   a    horizontal    line    how   many    horizontal 
planes  can  be  passed  ?  how  many  vertical  planes  ? 

10,  The  distance  from  a  point  A  to  a  point  13,  situated 
in  a  given  plane,  is  20  in.,  and  the  distance  from  B  to  the 
foot  of  the  perpendicular  dropped  from  A  to  the  plane  is 
16  in.  ;  find  the  distance  from  A  to  the  plane. 

Hi  The  distances  of  two  points,  A,  B,  from  a  given 
plane  are  5  in.,  11  in.,  respectively,  and  the  distance  be- 
tween the  feet  of  the  perpendiculars  dropped  from  A,  B 
to  the  plane  is  8  in.  ;  find  the  distance  from  A  to  B. 

12,  If  the   line   drawn   from   a  point  to  a  plane,  and 
making  an  angle  of  45°  with  the  plane,  is  10  in.  long,  what 
is  the  distance  from  the  point  to  the  plane  ? 

13,  Through  a  point  6  in.  from  a  plane  a  line  is  drawn 
to  the  plane,  making  the  angle  30°  with  the  plane  ;  find 
the  length  of  this  line. 

14,  Through  a  point  A  above  a  given  plane  a  perpen- 
dicular AO  is  drawn  to  the  plane;  about  0  as  centre  a 
circumference  is  described,  and  through  a  point  B  in  this 
circumference  a  tangent  B0\^>  drawn.    Given,  .4  0=12  in., 

15in.    find  AC. 


15,  A  line  makes,  with  a  plane,  an  angle  of  24°  ;  what 
are  the  least  and  the  greatest  angles  which  it  makes  with 
lines  drawn  through  its  foot  in  the  given  plane? 

16,  Describe   different   relative    positions   which    three 
planes  may  have. 

17,  The  line   AB  meets  three    parallel    planes   in  the 
points  A,  E,  B\  and  the  line  CD  meets  the  same  planes 
in  the  points  C,  F,  D.     If  AE=Q  in.,  £E=S  in.,  CD 
=  12  in.,  compute  CT'arid  FD. 


GEOMETRY.  3 


18.  What  is  the  locus  of  the  points  in  a  plane  which  are 
at  a  given  distance  from  a  given  point  not  in  the  plane  ? 

What  is  the  locus  in  space  of  all  points : 

19.  Equidistant  from  two  given  points? 

20.  Equidistant  from  three  given  points  ? 

21.  Equidistant  from  two  given  parallel  lines? 

22.  Equidistant  from  three  given  parallels  not  in  the 
same  plane  ? 

23.  Equidistant  from  two  given  intersecting  lines? 

24.  Equidistant   from   the   three  edges  of  a  trihedral 
angle  ? 

25.  At  a  given  distance  from  a  given  plane? 

26.  Equidistant  from  two  given  parallel  planes  ? 

27.  Equidistant  from  two  given  intersecting  planes  ? 

28.  Equidistant  from  the  three  faces  of  a  trihedral  angle  ? 

29.  What  is  the  locus  of  all  lines  drawn  through  a  given 
point,  and  parallel  to  a  given  plane  ? 

30.  What  is  the  locus  of  points  in  a  given  plane  which 
are  equidistant  from  two  given  points  not  in  the  plane  ? 

31.  What  is  the  locus  of  points  equidistant  from  two 
given  planes,  and  also  equidistant  from  two  given  points  ? 

32.  What  position  relative  to  a  given  plane  has  a  line 
if  its  projection  on  the  plane  is  equal  to  its  own  length? 
What  position  if  its  projection  is  a  point? 

33.  A  line  4  feet  long  makes,  with  a  plane,  the  angle 
45° ;  find  its  projection  on  the  plane. 

34.  If  the  projection  of  a  line  upon  a  plane  is  equal  to 
half  the  length  of  the  line,  what  is  the  angle  of  inclination 
of  the  line  to  the  plane  ? 


EXERCISE   MANUAL. 


§  2.   THE  PRISM. 

1,  One  edge  of  a  cube  =  a;  find  the  surface,  the  volume, 
and  the  length  of  a  diagonal. 

2,  Find  one  edge  and  the  volume  of  a  cube,  if  the  sur- 
face =  96  sq.  ft.     Also  find  the  same,  if  the  surface  =  /& 

3,  Find  the  edge  and  the  surface  of  a  cube,  if  the  vol- 
ume =  1000  cub.  ft.     Also  if  the  volume  =  V. 

4,  How  many  square  meters  of  surface  require  to  be 
cemented,  in  constructing  a  cubical  reservoir  which  will 
hold8000kgof  water? 

5,  Find  the  dimensions  of   a  cubical  block   of  marble 
which  weighs  2700kg.     Specific  gravity  of  marble  =  2.7. 

6,  Find  the  volume  of  a  cube,  if  a  diagonal  in  one  of 
its  faces  =  a. 

7,  How  much  lead  is  used  in  lining  the  bottom  and 
sides  of  a  cubical  vessel  that  holds  729  cub.  ft.  of  water  ? 

8,  If  a  cubical  vessel  requires  320  sq.  ft.  of  lead  for  lin- 
ing the  bottom  and  sides,  how  many  cubic  feet  of  water 
will  it  hold? 

9,  A  plumber  is  ordered  to  make  a  cubical  vessel  which 
will  hold  2  tons  of  water.     What  must  be  the  length  of 
one  edge  ? 

10,  A  cellar  12  ft.  long  and  9  ft.  wide  is  flooded  to.  a 
depth  of  4  in.     Find  the  weight  of  the  water. 

11,  What  is  the  weight  of  the  air  in  a  room  5m  long,  4m 
wide,  and  3.2m  high,  if  one  liter  of  air  weighs  1.293s? 

12,  How  much  lead  will  be  required  to  line  a  cistern, 
open  at  the  top,  which  is  4  ft.  6  in.  long,  2  ft.  8  in.  widet 
and  contains  42  cub.  ft.  ? 


GEOMETRY. 


13,  How  many  bricks  are  required  to  build  a  wall  90  ft. 
long,  8  ft.  high,  and  18  in.  thick,  if  the  bricks  are  9  in. 
long,  4-|-  in.  wide,  and  3  in.  thick  ? 

14,  A  book  is  8  in.  long,  6  in.  wide,  and  1J  in.  thick. 
Find  the  depth  of  the  box  whose  length  and  breadth  are 
3  ft.  4  in.  and  2  ft.   6  in.,  that  it  may  contain  400  such 
books. 

15,  An  open  cistern  is  made  of  iron  2  in.  thick.     The  in- 
ner dimensions  are  :  length,  4  ft.  6  in. ;  breadth,  3  ft. ;  depth, 
2  ft.  6  in.     What  will  the  cistern  weigh  (i.)  when  empty  ? 
(ii.)  when  full  of  water  ?     Specific  gravity  of  iron  =  7.2. 

16,  An  open  cistern  6  ft.  long  and  4^  ft.  wide  holds  108 
cub.  ft.  of  water.     How  many  cubic  feet  of  lead  will  it  take 
to  line  the  sides  and  bottom,  if  the  lead  is  ^  in.  thick? 

17,  Rain  has  fallen  to  the  depth  of  half  an  inch ;  how 
many  cubic  feet  of  water  have  fallen  on  an  acre  ? 

18,  The  three  dimensions  of  a  rectangular  parallelepiped 
are  a,  b,  c ;  find  the  surface,  the  volume,  and  the  length  of 
a  diagonal. 

19,  The  volume  of  a  parallelepiped  =  V,  and  the  three 
dimensions  are  as  m :  n :  p  ;  find  the  dimensions. 

Find  the  lateral  surface  and  the  volume  of  the  follow- 
ing regular  prisms,  if  in  each  case  the  height  =  h,  and  one 
side  of  the  base  =  a : 

20,  Triangular  prism.  22,    Hexagonal  prism. 

21,  Quadrangular  prism.  23,   Octagonal  prism, 

24.  Find  the  volume  of  a  right  triangular  prism,  if  its 
height  is  14  in.,  and  the  sides  of  the  base  are  6,  5,  and  5  in. 

25,  Find  the  volume  of  a  right  prism  14  ft.  high,  whose 
bases  are  squares,  each  side  measuring  1  ft.  6  in. 


6  EXERCISE   MANUAL. 

26.  The  base  of  a  right  prism  is  a  rhombus,  one  side  of 
which  =  10  in.,  and  the  shorter  diagonal  =  12  in.     The 
height  of  the  prism  —  15  in.     Find  the  entire  surface  and 
the  volume. 

27.  Find  the  volume  of  a  right  hexagonal  prism  whose 
height  is  10  ft.,  each  side  of  the  hexagon  being  10  in. 

28.  Find  the  volume  of  a  right  prism  32  ft.  long,  if  its 
ends  are  trapezoids,  the  parallel  sides  of  which  are  12  ft. 
and  8  ft.,  and  the  perpendicular  distance  between  them  is 
6ft. 

29.  Find  the  volume  of  a  right  prism,  if  the  height  =  3m, 
and  the  base  is  a  square  the  diagonal  of  which  =  2m. 

30.  How  many  cubic  inches  of  mahogany  will   be  re- 
quired to  veneer  the  top  of  a  table  in  the  shape  of  a  regu- 
lar hexagon,  each  side  of  which  measures  2  ft.,  the  veneer 
being  -|-  of  an  inch  thick  ? 

31.  How  many  cubic  feet  of  stone  are  required  to  build 
a  dam  1000  ft.  long,  20  ft.  high,  10  ft.  wide  at  the  bottom, 
and  4  ft.  wide  at  the  top  ? 

32.  The  wall  of  China  is  1500  miles  long,  20  ft.  high, 
15  ft.  wide  at  the  top,  and  25  ft.  wide  at  the  bottom ;  how 
many  cubic  yards  of  material  does  it  contain  ? 

33.  The  distance  around  a  reservoir  in  the  shape  of  a 
regular  hexagon  is  360  ft.     If  the  average  daily  loss  from 
evaporation  amounts  to  a  layer  of  water  2  in.  deep,  how 
many  cubic  feet  of  water  must  be  supplied  daily  to  main- 
tain the  water  at  a  constant  level  ? 

34.  Given  the  volume  Fand  the  height  A  of  a  regular 
hexagonal  prism,  find  one  side  a  of  the  base. 


GEOMETRY. 


§  3.   THE  CYLINDER  OF  REVOLUTION. 

1,  How  many  square  feet  of  sheet  iron  are  required  to 
make  a  funnel  2  ft.  in  diameter  and  40  ft.  long  ? 

2,  A   right   cylinder  is  10  ft.  high,  and  measures  7  ft. 
4  in.  around  the  base.     Find  the  convex  surface  and  the 
volume. 

3,  Find  the  radius  of  the  base  of  a  right  cylinder  if  the 
volume  =  1540  cu.  in.,  and  the  height  =  10  in. 

4,  Find  the  height  of  a  right  cylinder  if  the  volume  = 
3080  cub.  ft.,  and  the  radius  of  the  base  =  7  ft. 

5,  I  wish  to  have  made  a  cylindrical  pail  14  in.  high, 
and  holding  exactly  4  cub.  ft. ;  what  must  be  the  radius  of 
the  base  ? 

6,  If  the  total  surface  of  a  right  cylinder  closed  at  both 
ends  is  «,  and  the  radius  of  the  base  is  r,  what  is  the  height 
of  the  cylinder  ?  * 

7,  If  the  lateral  surface  of  a  right  cylinder  is  a,  and  the 
volume  is  b,  find  the  radius  of  the  base  and  the  height. 

8,  By  how  much   is   the  volume    of  a   right   cylinder 
increased  if  the  radius  of  the  base  is  doubled  ?  if  the  height 
is  doubled?  if  both  are  doubled? 

9,  What  will  it  cost  to  dig  a  well  3  ft.  in  diameter  and 
30  ft.  deep,  at  $4.00  per  cubic  yard  of  earth  thrown  out? 

10,  How  many  cubic  yards  of  earth  must  be  removed  in 
constructing  a  tunnel   100  yds.  long,  whose  section   is  a 
semicircle  with  a  radius  of  10  ft.  ? 

11,  If  the  diameter  of  a  well  is  7  ft.,  and  the  water  is 
10  ft.  deep,  how  many  gallons  of  water  are  there,  reckon- 
ing 7£  gals,  to  the  cubic  foot? 


8  EXERCISE    MANUAL. 


12.  If  a  cubic  foot  of  brass  is  drawn  into  a  wire  ^  of  an 
inch  in  diameter,  find  the  length  of  the  wire. 

13.  Find  the  weight  of  a  cylindrical  iron  shell  1  in.  thick 
and  2  ft.  long,  the  inner  radius  being  7  in.     Specific  grav- 
ity of  iron  —  7.2. 

14.  A  rectangular  sheet  of  tin   44  in.  long  and  14  in. 
wide  is  bent  so  as  to  form  a  cylindrical  surface  14  in.  high. 
Find  the  area  of  the  base,  and  the  volume. 

15.  Find  the  weight   of  mercury  contained   in   a   cyl- 
indrical vessel,  if  the  diameter  of  the  base  =  2dm  and  the 
depth  of  the  mercury  —  4dm.     Specific  gravity  of  mercury 

=  13.6. 

16.  A  cylindrical  vessel  holding  20  liters  has  a  height 
just  equal  to  the  diameter  of  the  base ;  find  its  dimensions. 

17,.  The  volume  of  a  right  cylinder  =  340  liters ;  find 
the  lateral  surface,  if  the  height  is  equal  to  twice  the  diam- 
eter of  the  base  ?  % 

18,  The  height  of  the  zinc  vessel  holding  1  liter  is  double 
the  diameter  of  the  base ;    the  thickness  of  the  metal  is 
0.5cm,  and  the  specific   gravity  of  zinc  is  7.19 ;  find  the 
weight  of  the  vessel. 

19,  How  many  cubic  feet  of  lead  will  be  required  to  line 
the  curved  surface  and  bottom  of  an  open  cylindrical  vessel 
whose  depth  is  7  ft.,  and  which  will  hold  198  cub,  ft.  of 
water,  if  the  lead  is  i  of  an  inch  thick  ? 

20,  A  marble   column  measures  7  ft.  4  in.  in  circumfer- 
ence, and  its  height  is  15  ft.  ;  find  the  expense  of  polishing 
the  curved  surface  at  $1.50  per  square  foot;  and  also  the 
additional  expense  if  the  whole  surface  is  polished. 

21,  A  cubic  foot  contains  7  J  gals.  If  375  gals,  are  pumped 
out  of  a  well   7  ft.  in  diameter,  how  many  inches  will  the 
surface  of  the  water  fall  in  consequence  ? 


GEOMETRY.  9 


22.  What  length  of  wire  0.08  of  an  inch  thick  can  be 
made  out  of  a  cubic  inch  of  metal  ? 

23.  How  many  revolutions  of  a  roller  3  ft.  in  length  and 
18  in.  in  diameter  would  it  take  to  go  over  a  grass  plot  half 
an  acre  in  extent? 

24.  About  the  convex  surface  of  a  right  cylinder  2  ft. 
4  in.  in  diameter,  and  3  ft.  4  in.  high,  a  cord  J  of  an  inch  in 
diameter  is  wound  until  the  surface  is  completely  covered. 
How  many  feet  of  cord  are  required  ? 

25.  A  cast  iron  cylinder  3  ft.  long  and  7  in.  in  diameter 
is  reduced  in  a  lathe  to  a  diameter  of  6  in. ;  find  the  loss  in 
weight.     Specific  gravity  of  iron  =  7.2. 

26.  The   piston   of  a  pump  is   16cm  in   diameter,  and 
moves  through  a  space  of  46cm.     How  many  liters  of  water 
are  thrown  out  by  1000  strokes  ? 

27.  When  a  body  is  placed  under  water  in  a  right  cylin- 
der 60cm  in  diameter,  the  level  of  the  water  rises  30cm ;  find 
the  volume  of  the  body. 

28.  How  much  will  a  brass  cylinder  weigh  under  water 
if  the  height  is  64cm,  and  the  diameter  of  the  base  is  40cm  ? 
Specific  gravity  of  brass  =  7.8.     The  loss  in  weight  of  an 
immersed  body  is  equal  to  the  weight  of  the  liquid  displaced. 

29.  If  the  circumference  of  the  base  of  a  right  cylinder 
is  c,  and  the  height  h,  find  the  volume  V. 

30.  Having  given  the  total  surface  2*  of  a  right  cylinder, 
in  which  the  height  is  equal  to  the  diameter  of  the  base, 
find  the  volume  V. 

31.  If  the  circumference  of  the  base  of  a  right  cylinder 
is  c,  and  the  total  surface  is  T,  find  the  volume  V, 


10  EXERCISE    MANUAL. 

32.  If  the  volume  of  a  right  cylinder  is  V,  and  height  h, 
find  the  total  surface  T. 

33.  If  the   volume    of   a  right   cylinder,  in  which   the 
height  is  equal  to  the  diameter  of  the  base,  is  V,  find  the 
height  h  and  the  total  surface  T. 

34.  Having  given  the  total  surface  Tof  a  right  cylinder, 
and  the  height  h,  find  the  diameter  and  the  volume. 


§  4.   THE  PYRAMID. 

1.  Two  pyramids  standing  on  the  same  plane  have  the 
same  height,  and  the  areas  of  their  bases  are  120  sq.  ft.  and 
180  sq.  ft.  respectively.     A  plane  parallel  to  the  plane  of 
their  bases  cuts  from  the  first  pyramid  a  section  of  70  sq.  ft. ; 
find  the  area  of  the  section  of  the  second  pyramid. 

2.  A  pyramid  15  ft.  high  has  a  base   containing   169 
sq.  ft.     At  what  distance  from  the  vertex  must  a  plane  be 
passed  parallel  to  the  base  so  that  the  section  may  contain 
100  sq.  ft.  ? 

3.  The  base  of  a  pyramid  contains  144  sq.  ft.     A  plane 
parallel  to  the  base  and  4  ft.  from  the  vertex  cuts  a  section 
containing  64  sq.  ft. ;  find  the  height  of  the  pyramid. 

4.  A  pyramid  12  ft.  high  has  a  square  base  measuring 
8  ft.  on  a  side.     What  will  be  the  area  of  a  section  made 
by  a  plane  parallel  to  the  base  and  4  ft.  from  the  vertex  ? 

5.  Two  pyramids  standing  on  the  same  plane  have  the 
same  height,  14  ft.     The  first  has  for  base  a  square  meas- 
uring 9  ft.  on  a  side ;  the  second  a  hexagon  measuring  7  ft. 
on  a  side.     What  will  be,  in  each  pyramid,  the  areas  of  the 
sections  made  by  a  plane  parallel  to  the  bases  and   6  ft. 
from  their  vertices? 


GEOMETRY.  11 


6.  The  base  of  a  regular  pyramid  is  a  hexagon  of  which 
the  side  measures  3  ft.     Find  the  height  of  the  pyramid  if 
the  lateral  area  is  equal  to  ten  times  the  area  of  the  base. 

Find  the  volume  in  cubic  feet  of  a  regular  pyramid : 

7.  When  its  base  is  a  square,  each  side  measuring  3  ft. 
4  in.,  and  its  height  is  9  ft. 

8.  When  its  base  is  an  equilateral  triangle,  each  side 
measuring  4  ft.,  and  its  height  is  15  ft. 

9.  When  its  base  is  a  regular  hexagon,  each  side  meas- 
uring 6  ft.,  and  its  height  is  30  ft. 

Find  the  total  surface  in  square  feet  of  a  regular  pyra- 
mid: 

10.  When  each  side  of  its  square  base  is  8  ft.,  and  the 
slant  height  is  20  ft. 

11.  When  each  side  of  its  triangular  base  is  6  ft.,  and 
the  slant  height  is  18  ft. 

12.  When  each  side  of  its  square  base  is  26  ft.,  and  the 
perpendicular  height  is  84  ft. 

Find  the  height  in  feet  of  a  regular  pyramid  when  its 
volume  and  its  base  are : 

13.  Volume  26  cub.  ft.  936  cub.  in.,  and  each  side  of  its 
square  base  3  ft.  6  in. 

14.  Volume  20  cub.  ft.,  and  the  sides  of  its  triangular 
base  5  ft.,  4  ft.,  and  3  ft. 

15.  The  base  edge  of  a  regular  pyramid  with  a  square 
base  measures  40  ft.,  the  lateral  edge  101  ft. ;  find  its  vol- 
ume in  cubic  feet. 


12  EXEECISE   MANUAL. 

16.  The  volume  of   a  regular  pyramid  with   a   square 
base  is  200  cub.  ft.,  its  base  edge  5  ft. ;  find  the  total  surface 
in  square  feet. 

17.  The  base  edge  of  a  regular  triangular  pyramid  is 
18cm,  the   lateral   edge   15cm ;    find   the  slant   height,  the 
area  of  a  lateral  face,  the  height  of  the  pyramid,  the  total 
surface,  and  the  volume. 

18.  How  many  square  feet  of  ground  will  the  base  of  a 
pyramid  cover  which  contains  102  cub.  yds.,  21  cub.  ft.,  and 
the  height  of  which  is  45  ft.  ? 

19.  A   cubic  inch  of  gold  weighs   11  oz.  Avoir. ;   how 
many  ounces  will  be  used  in  making  a  solid  gold  ornament 
in  the  shape  of  a  regular  pyramid  with  a  square  base,  if 
each  side  of  its  base  is  3  in.  and  its  height  6  in.  ? 

20.  Find  the  volume  of  a  regular  pyramid  whose  slant 
height  is  12  ft.  and  whose  base  is  an  equilateral  triangle 
inscribed  in  a  circle  having  a  radius  of  10  ft. 

21.  Having  given  the  base  edge  a,  and  the   total  sur- 
face T,  of  a  regular  pyramid  with  a  square  base,  find  the 
volume  V. 

22.  The  base  edge  of  a  regular  pyramid  whose  base  is  a 
square  is  a,  the  total  surface  T;  find  the  height  of  the 
pyramid. 

23.  The  eight  edges  of  a  regular  pyramid  with  a  square 
base  are  equal  in  length,  and  the  total  surface  is  T;  find  the 
length  of  one  edge. 

24.  Find  the  lateral  edge  of  a  regular  pyramid  with  a 
square  base,  having  given  the  height  h  and  the  total  sur- 
face T. 


GEOMETRY.  13 


§  5.   THE  CONE  OF  REVOLUTION. 

1.  The  height  of  a  right  cone  is  2m,  and  the  area  of  its 
base  lqm ;  find  the  area  of  the  section  made  by  a  plane  par- 
allel to  the  base,  and  80cm  from  the  vertex. 

2.  The  radius  of  the  base  of  a  right  cone  is  40  in.,  and 
the  height  of  the  cone  is  8  ft.     At  what  distance  from  the 
vertex  must  a  plane  be  passed  parallel  to  the  base  that  the 
section  may  be  a  circle  with  a  radius  of  30  in.  ? 

3.  If  a  right  cone  is  4  ft.  high,  at  what  distance  from 
the  vertex  must  a  plane  be  passed  parallel  to  the  base,  in 
order  that  the  section  may  be  equal  to  ^  of  the  base? 

4.  If  the  slant  height  of  a  right  cone  is  a,  and  the  area 
of  the  base  5,  find  the  area  of  a  section  made  by  a  plane 
parallel  to  the  base,  at  the  distance  c  from  the  vertex. 

5.  The  slant  height  of  a  right  cone  is  2  ft.     At  what 
distance  from  the  vertex  must  the  slant  height  be  cut  by  a 
plane  parallel  to  the  base,  in  order  that  the  lateral  surface 
may  be  divided  into  two  equivalent  parts? 

6.  The  height  of  a  right  cone  is  equal  to  the  diameter 
of  its  base ;  find  the  ratio  of  the  area  of  the  base  to  the 
lateral  surface. 

7.  The  area  of  the  convex  surface  of  a  right  cone  is  99 
sq.  ft.,  and  the  radius  of  its  base  is  2  ft.  7£  in. ;  find  its 
slant  height. 

8.  The  slant  height  of  a  right  cone,  whose  convex  sur- 
face measures  2310  sq.  ft.  is  35  ft. ;  find  its  height. 

9.  How  many  cubic  feet  of  lead  will  be  required  to 
cover   a  conical  spire  which   measures  35  ft.    round   the 
base,  and  has  a  slant  height  of  30  ft.,  if  the  lead  is  -f%  of 
an  inch  thick  ? 


14  EXERCISE   MANUAL. 

10.  Find  the  expense  of  covering  a  conical  spire  which 
measures  40  ft.  round  the  base,  and  has  a  slant  height  of 
30  ft.,  with  lead  -j-  of  an  inch  thick,  if  one  cubic  inch  of 
lead  weighs  6z  oz.,  and  lead  is  worth  8  cents  a  pound. 

11.  How  much  canvas  is  required  for  a  conical  tent,  the 
altitude  of  which  is  8  ft.,  and  the  diameter  of  the  base  is 
7ft.? 

12.  A  conical  tent,  whose  slant  height  is  12  ft.,  requires 
132  sq.  ft.  of  canvas  to  cover  it ;  find  how  much  ground 
the  tent  covers. 

13.  What  will  be  the  cost  of  painting  a  conical  spire 
whose  slant  height  is  118  ft.,  and  whose  circumference  at 
the  base  is  64  ft.,  at  16  cents  a  square  yard? 

14.  What  length  of  canvas  f  of  a  yard  wide  is  required 
to  make  a  conical  tent  12  ft?,  in  diameter  and  8  ft.  high  ? 

15.  The  slant  height  of  a  right  cone  is  equal  to   the 
diameter  of  the  base.     If  the  slant  height  is  a,  find  the 
total  surface. 

16.  What  does  the  volume  V  of  a  cone  become,  if  the 
height  is  doubled  ?     If  the  radius  of  the  base  is  doubled  ? 
If  both  the  height  and  the  radius  of  base  are  doubled  ? 

17.  A  conical  mound  of  earth  measures  264  yds.  round 
the  base,  and  the  length  of  its  slope  is  70  yds. ;  find  the 
number  of  cubic  yards  in  the  mound. 

18.  Find  the  height  of  a  conical  vessel  whose  diameter 
at  the  base  is  2  ft.  4  in.,  in  order  that  its  volume  may  be 
27  cub.  ft. 

19.  Find  the  volume  of  a  right  cone,  the  height  of  which 
is  15  ft.,  and  the  circumference  of  the  base  14  ft. 

20.  The  circumference  of  the  base  of  a  cone  is  12J  ft., 
and  its  height  8J  ft. ;  find  its  volume. 


GEOMETRY.  15 


21.  A  right  oone,  3  ft.  high  and  2  ft.  in  diameter  at  the 
base,  is  placed  on  the  ground,  and  sand  is  poured  over  it 
until  it  makes  a  conical  heap  5  ft.  high  and  30  ft.  in  cir- 
cumference ;  how  many  cubic  feet  of  sand  are  used  ? 

22.  How  many  gallons  are  contained  in  a  conical  vessel 
the  radius  of  whose  base  is  6  ft.,  and  the  slant  height  10 
feet?     (A  gallon  is  231  cub.  in.) 

23.  A  conical  wine-glass  is  2  in.  wide  at  the  top,  and  3 
in.  deep  ;  how  many  cubic  inches  of  wine  will  it  hold  ? 

24.  The  slant  height  of  a  right  cone  is  5  ft.     A  plane 
parallel  to  the  base  cuts  the  slant  height  at  a  distance  of 
2  ft.  from  the  vertex.     If  the  section  made  by  this  plane  is  a 
circle  whose  radius  is  4J-  in.,  find  the  volume  of  the  cone. 

25.  A  right  cone  made  of  silver,  whose  height  is  twice 
the  diameter  of  its  base,  weighs  2.5kg ;  required  the  dimen- 
sions of  the  cone,  the  specific  gravity  of  silver  being  10.47. 

26.  A  right  cone  is  4  ft.  high.      The  section  made  by  a 
plane  parallel  to  the  base  at  the  distance  of  1  ft.  from  the 
vertex  contains  1  sq.  ft.  ;  find  the  volume  of  the  cone. 

27.  The  volume  of  a  right  cone  is  -J  of  a  cubic  meter, 
the  radius  of  the  base  is  lm  ;  find  the  lateral  surface. 

When  the  slant  height  of  a  right  cone  is  equal  to  the 
diameter  of  its  base  : 

28.  Find  the  diameter,  if  the  volume  is  25ccm. 

29.  Find  the  volume,  if  the  total  surface  is  1000qcm. 

30.  Find  the  volume,  if  the  total  surface  is  T. 

31.  Given  the  total  surface  T  of  a  right  cone,  and  the 
radius  r  of  the  base  ;  find  the  volume  V. 

32.  Given  the  total  surface   T  of  &  right  cone,  and  the 
lateral  surface  $;  find  the  volume  V. 


16  EXERCISE    MANUAL. 


§6.   FRUSTUMS  OF  PYRAMIDS  AND  CONES. 

1.  How  many  square  feet  of  tin  will    be    required    to 
make  a  funnel  if  the  diameters  of  the  top  and  bottom  are  to 
be  28  in.  and  14  in.  respectively,  and  the  height  24  in.  ? 

2.  The  slant  height  of  a  church  spire  in  the  shape  of  the 
frustum  of  a  regular  hexagonal  pyramid  is  20  ft.,  the  length 
of  each  side  of  its  base  is  5  ft.,  and  of  its  top  2  ft.     How 
many  square  feet  of  lead  will  be  required  to  cover  its  lateral 
faces  and  top  ? 

3.  Find  the  expense  of  polishing  the  curved  surface  of  a 
marble  column  in  the  shape  of  the  frustum  of  a  right  cone 
whose  slant  height  is  12  ft.,  and  the  radii  of  the  circular 
ends  are  3  ft.  6  in.  and  2  ft.  4  in.  respectively,  at  60  cents 
per  square  foot. 

4.  Find  the  lateral  surface  of  the  frustum  of  a  right  cone 
if  the  radii  of  the  bases  are  2m  and  6m,  and  the  height  is  3m. 

5.  Having  given  the  lateral  surface  of  the  frustum  of  a 
right   cone,  3454qdm,  and  the  radii  of  the  bases,  1.42m  and 
0.64m,  find  the  height  of  the  frustum. 

6.  The  height  of  a  right  cone  is  15  ft.,  the  radius  of  its 
base  2  ft.     Find  the  lateral  surface  of  the  frustum  made  by 
a  plane  parallel  to  the  base,  and  distant  12  ft.  from  the 
vertex. 

7.  Having  given  the  slant  height  a  of  a  right  cone,  and 
the  radius  r  of  the  base,  find  at  what  distance  from  the 
vertex  the  slant  height  must  be  cut  by  a  plane  parallel  to 
the  base  in  order  that  the  total  surfaces  of  the  two  parts 
into  which  the  cone  is  divided  may  be  equal. 

8.  A  round  stick  of  timber  is  20  ft.  long,  3  ft.  in  diame- 
ter at  one  end,  2.6  ft.  at  the  other ;  how  many  cubic  feet 
does  it  contain  ? 


GEOMETRY.  17 


9.  Find  the  volume  of  the  frustum  of  a  right  cone ;  given 
the  radii  of  the  circular  ends,  3  ft.  and  3  ft.  10  in.,  and  the 
slant  height,  2  ft.  2  in. 

10.  Find  the  volume  of  the  frustum  of  a  regular  square 
pyramid  whose  height  is  24  ft.,  and  the  sides  of  its  square 
ends  are  respectively  9  ft.  and  4  ft. 

11.  The  slant  height  of  the  frustum  of  a  regular  square 
pyramid  is  20  ft.,  the  length  of  each  side  of  its  base  40  ft., 
of  each  side  of  its  top  16  ft. ;  find  its  volume. 

12.  The  mast  of  a  ship  is  51  ft.  high,  and  the  circum- 
ferences of  its  ends  are  5  ft.  6  in.  and  1  ft.  10  in.     What 
will  it  cost  at  60  cents  per  cubic  foot  ? 

13.  A  bucket  is  16  in.  deep,  18  in.  wide  at  the  top,  and 
12  in.  wide  at  the  bottom.     How  many  gallons  of  water 
will  it  hold,  reckoning  7£  gals,  to  the  cubic  foot  ? 

14.  The  frustum  of  a  regular  pyramid  is  5  ft.  high,  and 
its    bases  are  two  regular  octagons  whose  sides  are  4  ft. 
and  3  ft.  respectively.     Find  the  volume  of  the  frustum. 

15.  If  the  bases  of  the  frustum  of  a  pyramid  are  two 
regular  hexagons  whose  sides  are  1   ft.  and  2  ft.  respec- 
tively, and  the  volume  of  the  frustum  is  12  cu.  ft.,  find  its 
height. 

16.  A  plane  parallel  to  the  base  of  a  right  cone  cuts 
from  the  cone  a  frustum  whose  volume  is  20cbm ,  and  the 
radii  of  the  two  bases  are  3m  and  2m  respectively.     Find 
the  volume  of  the  entire  cone. 

17.  The  radii  of  the  bases  of  the  frustum  of  a  cone  are 
3.5m  and  7.3m,  and  the  height  of  the  frustum  is  2m  ;  find  the 
surface  and  the  volume  of  the  entire  cone. 


18  EXERCISE    MANUAL. 


18.  The  height  of  a  right  cone  is  10m,  the  radius  of  its 
base  5m.     What  must  be  the  distance  from  the  base  of  a 
plane  parallel  to  the  base  in  order  that  the  volume  of  the 
frustum  made  by  the  plane  may  be  20cbm? 

19.  The  frustum  of  a  right  cone  is  14  ft.  high,  and  has  a 
volume  of  924  cub.  ft.     Find  the  radii  of  its  bases  if  their 
sum  is  9  ft. 

20.  From  a  right  cone  whose  slant  height  is  30  ft.,  and 
circumference  of  whose  base  is  10  ft.,  there  is  cut  off  by  a 
plane  parallel  to  the  base  a  cone  whose  slant  height  is  6  ft. 
Find  the  convex  surface  and  the  volume  of  the  frustum. 

21.  The  interior  dimensions  of  an  iron  vessel,  open  at  the 
top  and  in  the  shape  of  the  frustum  of  a  right  cone,  are : 
diameter  at  the  top  1  ft.  9  in.,  at  the  bottom  7  in.,  depth 

1  ft.  8  in. ;  and  the  corresponding  exterior  dimensions  are 

2  ft.,  9  in.,  and  1  ft.  10  in.     How  many  cubic  inches  of  .iron 
were  used  in  its  construction  ? 

22.  A  Dutch  windmill  in  the  shape  of  the  frustum  of  a 
right  cone  is  12m  high.     The  outer  diameters  at  the  bottom 
and  the  top  are  16m  and  12m,  the  inner  diameters  12m  and 
10m,  respectively.     How  many  cubic  meters  of  stone  were 
required  to  build  it? 

23.  The  chimney  of  a  factory  has  the  shape  of  the  frus- 
tum of  a  regular  pyramid.     Its  height  is  180  ft.,  and  its 
upper  and  lower  bases  are  squares  whose  sides  are  16  ft. 
and  10  ft.,  respectively.     The  section  of  the  flue  is  through- 
out a  square  whose  side  is  7  ft.     How  many  cubic  feet  of 
material  does  the  chimney  contain? 

24.  Find  the  volume  V  of  the  frustum  of  a  cone  of  rev- 
olution, having  given  the  slant  height  a,  the  height  h,  and 
the  convex  surface  S. 


GEOMETRY.  19 


§7.   THE  SPHERE. 

1.  What  is  the  locus  of  the  tangents  which  can  be  drawn 
from  an  exterior  point  to  a  given  sphere  ? 

2.  The  radius  of  a  sphere  is  2m.     Find  the  area  of  a  sec- 
tion made  by  a  plane,  the  distance  of  which  from  the  centre 
of  the  sphere  is  40cm. 

3.  What  is  the  distance  of  a  small  circle  the  area  of 
which  is  3qm,  from  the  centre  of  a  sphere  the   radius   of 
which  is  2m  ? 

4.  The  radius  of  a  sphere  is  7  ft.     Find  the  distance 
from  the  pole  of  a  great  circle  to  its  circumference,  (i.)  when 
measured  on  the  surface  of  the  sphere ;  (ii.)  when  measured 
in  a  straight  line. 

5.  The  distance  from  the  two  poles  of  a  small  circle 
drawn  on  the  surface  of  a  sphere  to  the  circumference  of 
the  circle  are  3m  and  4m  respectively.     Find  the  area  of  this 
circle. 

6.  In  order  to  find  the  radius  of  a  sphere,  describe  a 
small   circle  with   a   pair    of   dividers,  and   measure   the 
length  of  its  circumference  with  the  help  of  a  string.     If 
the  opening  of  the  dividers  is  6  in.,  and   the   length   of 
the  circumference  is  22  in.,  find  the  radius  of  the  sphere. 

7.  Find  the  surface  of  a  sphere  if  the  diameter  is  (i.) 
10  in. ;  (ii.)  1  ft.  9  in. ;  (iii.)  2  ft.  4  in. ;  (iv.)  7  ft. ;  (v.) 
4.2  ft.  •  (vi.)  10.5  ft. 

8.  Find  the  diameter  of  a  sphere  if  the  surface  is  (i.) 
616  sq.  in.  ;  (ii.)  38£  sq.  ft. ;  (iii.)  9856  sq.  ft. 

9.  The  circumference  of  a  dome  in  the  shape  of  a  hemis- 
phere is  66  ft. ;  how  many  square  feet  of  lead  are  required 
to  cover  it? 


20  EXEKCISE    MANUAL. 


10.  How  many  square  feet  of  lead  are  required  to  make 
1C  hemispherical  bowls,  if  the  diameter  of  each  is  2  ft.  4  in.  ? 

11.  If  the  ball  on  the  top  of  St.  Paul's  Cathedral  in 
London  is  6  ft.  in  diameter,  what  would  it  cost  to  gild  it  at 
7  cents  per  square  inch  ? 

12.  If  the  surface  of  a  sphere  is  JS,  what  is  the  surface  of 
a  sphere  having  a  radius  twice  as  large  ? 

13.  If  the  surface  of  a  sphere  is  jS,  find  the  circumference 
0  of  a  great  circle. 

14.  What  is   the  numerical  value  of  the   radius  of  a 
sphere  if  its  surface  has  the  same  numerical  value  as  the 
circumference  of  a  great  circle  ? 

15.  Find  the  surface  of  a  lune  if  its  angle  is  30°,  and  the 
total  surface  of  the  sphere  is  4  sq.  ft. 

16.  What  is  the  angle  of  a  lune  if  its  surface  is  1  sq.  ft., 
and  the  total  surface  of  the  sphere  is  9  sq.  ft.  ? 

17.  What  fractional  part  of  the  whole  surface  of  a  sphere 
is  a  spherical  triangle  whose  angles  are  43°  27',  81°  57',  and 
114°  36'? 

18.  The  angles  of  a  spherical  triangle  are  60°,  70°,  and 
80°.     The  radius  of  the  sphere  is  14  ft.     Find  the  area  of 
the  triangle  in  square  feet. 

19.  The  sides  of  a  spherical  triangle  are  38°,  74°,  and 
128°.     The  radius  of  the  sphere  is  14  ft.     Find  the  area  of 
the  polar  triangle  in  square  feet. 

20.  The  sides  of  a  spherical  triangle  are  each  equal  to 
90°.    What  part  of  the  whole  surface  of  the  sphere  does  the 
triangle  contain? 


GEOMETRY.  21 


21.  The  radius  of  a  sphere  is  7  ft.     Find  the  area  of  a 
spherical  triangle  formed  by  the  arcs  of  three  great  circles, 
two  of  which  are  perpendicular  to  the  third,  and  make  with 
each  other  an  angle  of  60°. 

22.  Through   a  diameter  of  a  sphere   two   planes  are 
passed  so  as  to  cut  from  the  circumference  of  a  great  circle, 
perpendicular  to  both  the  planes,  an  arc  of  45°.    The  area 
of  the  spherical  surface  included  by  these  planes  is   16  TT 
sq.  ft.     Find  the  radius  of  the  sphere. 

23.  On  a  sphere  whose  radius  is  21  ft.,  find  the  area  of  a 
spherical*triangle  polar  to  one  whose  perimeter  is  equal  to 
half  the  circumference  of  a  great  circle. 

24.  Find  the  area  of  a  spherical  polygon  on  a  sphere 
whose  radius  is  10|  ft.,  if  its  angles  are  100°,  120°,  140°, 
and  160°. 

25.  The  planes  of  the  faces  of  a  quadrangular  spherical 
pyramid  make  with  each  other  angles  of  80°,  100°,  120°,  and 
150° ;  and  the  length  of  a  lateral  edge  of  the  pyramid  is 
42  ft.     Find  the  area  of  its  base  in  square  feet. 

26.  The  planes  of  the  faces  of  a  triangular  spherical  pyr- 
amid make  with  each  other  angles  of  40°,  60°,  and  100°,  and 
the  area  of  the  base   of  the  pyramid  is  4?r  sq.  ft.     Find 
the  radius  of  the  sphere. 

27.  The  diameter  of  a  sphere  is  21  ft.     Find  the  curved 
surface  of  a  segment  whose  height  is  5  ft. 

28.  What  is  the  area  of  a  zone  of  one  base  whose  height 
is  h,  and  the  radius  of  the  base  r?    What  would  be  the 
area  if  the  height  were  twice  as  great  ? 

29.  In  a  sphere  whose  radius  is  r}  find  the  height  of  a 
zone  whose  area  is  equal  to  that  of  a  great  circle. 


22  EXERCISE   MANUAL. 

30.  The  radius  of  a  sphere  is  4  ft.     The  sphere  is  cut  by 
two  parallel  planes  situated  on  the  same  side  of  the  centre, 
at  the  distances  2  ft.  and  3  ft.  respectively  from  the  centre. 
Find  the  area  of  the  zone  formed  by  the  planes,  and  the 
area  of  each  of  the  circles  which  form  its  bases. 

31.  If  the  radius  of  a  sphere  is  r,  and  the  height  of  a 
zone  on  the  sphere  is  A,  find  the  radius  of  a  circle  equiva- 
lent to  the  area  of  the  zone4 

32.  How  many  square  feet  of  lead  will  be  required  to 
line   the  inside  of  a  bowl  in  the   shape  of  a  segment  of  a 
sphere  which  measures  40  in.  across  the  top,  and  whose 
greatest  depth  is  10  in.  ? 

33.  Find  the  convex  surface  of  a  slice  2  ft.  high,  cut  from 
a  spliere  whose  radius  is  17  ft. 

34.  The  height  of  a  spherical  zone  is  8  ft. ;  its  bases  are 
on  opposite  sides  of  the   centre,  and  their  radii  are  10  ft. 
and  6  ft.     Find  the  area  of  the  zone. 

35.  The  altitude  of  the  torrid  zone  is  about  3200  miles. 
Find  its  area  in  square  miles,  assuming  the  earth  to  be  a 
sphere  with  a  radius  of  4000  miles. 

36.  A  plane  divides  the  surface  of  a  sphere  of  radius  r 
into  two  zones,  such  that  the  surface  of  the  greater  is  a 
mean  proportional  between  the  entire  surface  and  the  sur- 
face of  the  smaller.     Find  the  distance  of  the  plane  from 
the  centre  of  the  sphere. 

37.  If  a  sphere  of  radius  r  is  cut  by  two  planes  equally 
distant  from  the  centre,  so  that  the  area  of  the  zone  com- 
prised between  the  planes  is  equal  to  the  sum  of  the  areas 
of  its  bases,  find  the  distance  of  either  plane  from  the  centre. 

38.  Find  the  area  of  the  zone  generated  by  an  arc  of  30°, 
of  which  the  radius  is  r,  and  which  turns  around  a  diam- 
eter passing  through  one  of  its  extremities. 


GEOMETRY.  23 


39.  Find  the  area  of  the  zone  of  a  sphere  of  radius  r, 
illuminated  by  a  lamp  placed  at  the  distance  Ti  from  the 
sphere. 

40.  How  much  of  the  earth's  surface  would  a  man  see  if 
he  were  raised  to  the  height  of  the  radius  above  it? 

41.  To  what  height  must  a  man  be  raised  above  the  earth 
in  order  that  he  may  see  one-sixth  of  its  surface  ? 

42.  Two  cities   are  200  miles  apart.     To  what  height 
must  a  man  ascend  from  one  city  in  order  that  he  may  see 
the  other,  supposing  the  circumference  of  the  earth  to  be 
25,000  miles? 

43.  Find  the  volume  of  a  sphere  if  the  diameter  is  (i.) 
14  in. ;  (ii.)  3  ft.  6  in. ;  (iii.)  10  ft.  6  in. ;  (iv.)  17  ft.  6  in. ; 
(v.)  14.7  ft. ;  (vi.)  42  ft. 

44.  Find  the  diameter  of  a  sphere  if  the  volume  is  (i.) 
75  cub.  ft.  1377  cub.  in.;  (ii.)  179  cub.  ft.  1152  cub.  in.; 
(iii.)  1047.816  cub.  ft. ;  (iv.)  38.808  cub.  yds. 

45.  Find  the  volume  of  a  sphere  whose  circumference  is 
45ft. 

46.  Find  the  volume  F"of  a  sphere  in  terms  of  the  cir- 
cumference (7  of  a  great  circle. 

47.  Find  the   radius  r  of  a  sphere,  having  given  the 
volume  V. 

48.  Find  the  radius  r  of  a  sphere,  if  its  circumference 
and  its  volume  have  the  same  numerical  value. 

49.  Find  the  volume  of  a  sphere  if  the  section  made  by 
a  plane  0.2m  from  the  centre  contains  0.8qm. 

50.  How  many  cubic  inches  of  lead  are  required  to  make 
a  spherical  shell  \  of  an  inch  thick,  if  the  exterior  diamctor 
is  to  be  7  in.  ? 


24  EXERCISE    MANUAL, 


51.  How  many  gallons  of  water  will  a   hemispherical 
bowl  hold,  whose  diameter  is  21  in.,  if  1  cub.  ft.  contains 
7*  gals.  ? 

52.  Find  the  weight  of  a  solid  ball  of  iron  whose  diame- 
ter is  5  in.,  if  1  cub.  in.  of  iron  weighs  4z  oz. 

53.  What  must  be  the  exterior  diameter  of  a  spherical 
shell  2  of  an  inch  thick,  that  its  hollow  part  may  contain 
113|cub.in.? 

54.  If  the  earth's  diameter  be  8000  miles,  and  geologists 
knew  the  interior  to  the  depth  of  5  miles  below  the  surface, 
what  fraction  of  the  whole  volume  would  be  known  ? 

55.  Find  the  diameter  of  the  mouth  of  a  cannon  that 
carries  a  ball  weighing  24   Ibs.,  when  1  cub.  in.  of  iron 
weighs  4 £  oz. 

56.  What  is  the  weight  of  a  spherical  shell  10  in.  in 
diameter  and  2  in.  thick,  composed  of  a  substance  of  which 
1  cub.  ft.  weighs  216  Ibs.  ? 

57.  The  weight  of  an  iron  ball  whose  diameter  is  4  in. 
is  9  Ibs. ;  find  the  weight  of  a  shell  whose  external  and 
internal  diameters  are  8  in.  and  5  in.  respectively. 

58.  If  an  iron  ball  4  in.  in  diameter  weighs  9  Ibs.,  what 
is  the  weight  of  a  hollow  iron  shell  2  in.  thick,  whose  exter- 
nal diameter  is  20  in.  ? 

59.  The  radius  of  a  sphere  is  7  ft. ;  what  is  the  volume 
of  a  wedge  whose  angle  is  36°  ? 

60.  What  is  the  angle  of  a  spherical  wedge,  if  its  vol- 
ume is  1  cub.  ft.,  and  the  volume  of  the  entire  sphere  is 
6  cub.  ft.  ? 

61.  What  is  the  volume  of  a  spherical  sector,  if  the 
area  of  the  zone  which  forms  its  base  is  3  sq.  ft.,  and  the 
radius  of  the  sphere  is  1  ft.  ? 


GEOMETRY.  25 


62.  The  volume  of  a  spherical  sector  is  0.48cbm,  and  the 
radius  of  the  sphere  is  2m ;  find  the  area  of  the  base  of  the 
sector  ? 

63.  If  the  volume  of  a  spherical  sector  is  a,  and  the 
area  of  its  base  b,  what  is  the  volume  V  of  the  sphere  to 
which  the  sector  belongs? 

64.  Find  the  volume  of  a  triangular  spherical  pyramid, 
if  the  angles  of  the  spherical  triangle  which  forms  its  base 
are  each  100°,  and  the  radius  of  the  sphere  is  7  ft.? 

65.  The  circumference  of  a  sphere  is  28  TT  ft.  ;  find  the 
volume  of  that  part  of  the  sphere  included  by  the  faces  of 
a  trihedral  angle  at  the  centre,  the  face  angles  of  which 
are  80°,  105°,  and  140°. 

66.  The  planes  of  the  faces  of  a  quadrangular  spherical 
pyramid  make  with  each  other  angles  of  80°,  100°,  120°, 
and  150°,  and  a  lateral  edge  of  the  pyramid  is  3-J-  ft. ;  find 
the  volume  of  the  pyramid. 

67.  The  radius  of  the  base  of  the  segment  of  a  sphere 
is  40  ft.,  and  its  height  is  20  ft. ;  find  its  volume. 

68.  The  radius  of  the  base  of  the  segment  of  a  sphere 
is  16  in.,  and  the  radius  of  the  sphere  is  20  in. ;  find  its 
volume. 

69.  The  inside  of  a  wash-basin  is  in  the  shape  of  the 
segment  of  a  sphere;  the  distance  across  the  top  is  16  in., 
and  its  greatest  depth  is  6  in. ;  find  how  many  pints  of 
water  it  will  hold,  reckoning  7-J  gallons  to  the  cubic  foot. 

70.  What  is  the  height  of  a  zone,  if  its  area  is  S,  and 
the  volume  of  the  sphere  to  which  it  belongs  is  V? 

71.  The  radii  of  the  bases  of  a  spherical  segment  are 
6  ft.  and  8  ft.,  and  its  height  is  3  ft. ;  find  its  volume. 


26  EXERCISE   MANUAL. 

72.  Find  the   volume  of  a   spherical   segment,  if   the 
diameter  of  each  base  is  8  ft.,  and  the  height  is  4  ft. 

73.  A  bowl  has  the  shape  of  a  spherical  segment  of  two 
bases.     It  measures  2  ft.  across  the  top,  and  10  in.  across 
the  bottom,  and  its  greatest  depth  is  13  in. ;  how  many  gal- 
lons of  water  will  it  hold,  if  1  cub.  ft.  contains  7-^-  gallons? 

74.  A  sphere  whose  radius  is  7  ft.  is  cut  by  two  parallel 
planes  on  the  same  side  of  the  centre,  at  the  distances  from 
the  centre  of  3  ft.  and  5  ft.  respectively ;  find  the  volume 
of  the  spherical  segment  comprised  between  the  planes. 

75.  The  radius  of  a  sphere  is  10m ;  find  the  volume  of  the 
segment  cut  from  the   sphere  by  a  plane  whose  distance 
from  the  centre  is  equal  to  half  the  radius. 

76.  A  sphere  is  cut  by  a  plane  whose  distance  from  the 
centre  is  equal  to  half  the  radius ;  find  the  ratio  of  the  two 
parts  into  which  this  plane  divides  (i.)  the  surface  of  the 
sphere;  (ii.)  the  volume  of  the  sphere. 

77.  A  spherical  segment  of  one  base  is  half  as  large  as 
the  spherical  sector  to  which  it  belongs  ;  having  given  the 
radius  r  of  the  sphere,  find  the  height  of  the  segment. 

78.  The  radii  of  two  concentric  spheres  are  r,  r';  find 
the  volume  of  the  portion  contained  between  two  parallel 
planes  passing  on  the  same  side  of  the  common  centre  at 
the  distances  a  and  a-\-b  from  the  centre. 

79.  Having  given  the  volume  V,  and  the  height  h,  of  a 
spherical  segment  of   one  base,  find  the  radius  r  of  the 
sphere. 

80.  Find  the  weight  of   a  sphere  of  radius  r,  which 
floats  in  a  liquid  of  specific  gravity  s,  with  i  of  its  surface 
above  the  surface  of  the  liquid.     The  weight,  of  a  floating 
body  is  equal  to  the  weight  of  the  liquid  displaced. 


GEOMETRY.  27 


§  8.   EQUIVALENT  SOLIDS. 

1.  What  is  the  ratio  of  the  heights  of  two  equivalent 
right  prisms,  if  the  areas  of  their  bases  are  as  4 : 5  ? 

2.  A  cube  whose  edge  is  12cm  long  is  transformed  into  a 
right  prism  whose  base  is  a  rectangle  16cm  long  and  12cni 
wide;    find  the  height  of  the  prism,   and  the   difference 
between  its  total  surface  and  the  surface  of  the  cube. 

3.  The  dimensions  of  a  rectangular  parallelepiped  are 
a,  5,  c.     Find  (i.)  the  height  of  an  equivalent  right  cylin- 
der having  a  for  the  radius  of  its  base;  (ii.)'the  height  of 
an  equivalent  right  cone  having  a  for  the  radius  of  its  base  ; 
(iii.)  the  radius  of  an  equivalent  sphere. 

4.  A  cube  of  lead  whose  edge  is  a  is  melted  and  recast 
in  the  shape  of  a  sphere ;  find  the  radius  of  the  sphere. 

5.  A  sphere  of  lead  whose  radius  is  r  is  melted  and  re- 
cast in  the  shape  of  a  cube ;  find  the  length  of  one  edge  of 
the  cube. 

6.  A  regular  pyramid  12  ft.  high  is  transformed  into 
a  regular  prism  with  an  equivalent   base;    what  is  the 
height  of  the  prism? 

7.  The  diameter  of  a  cylinder  is  14  ft.,  and  its  height  is 
8  ft. ;  find  the  height  of  an  equivalent  right  prism,   the 
base  of  which  is  a  square  with  a  side  4  ft.  long. 

8.  If  one  edge  of  a  cube  is  a,  what  is  the  height  h  of 
an  equivalent  right  cylinder  whose  diameter  is  b  ? 

9.  The  diameters  of  two  equivalent  cylinders  are  as  1 :  4. 
The  height  of  the  first  is  48  ft. ;  what  is  the  height  of  the 
other? 


28  EXERCISE    MANUAL. 

10.  The  heights  of  two  equivalent  right  cylinders  are 
as  4  :  9.     The  diameter  of  the  first  is  6  ft. ;  what  is  the 
diameter  of  the  other? 

11.  A  right  cylinder  6  ft.  in  diameter  is  equivalent  to  a 
right  cone  7  ft.  in  diameter.     If  the  height  of  the  cone  is 
8  ft.,  what  is  the  height  of  the  cylinder? 

12.  A  sphere  whose  diameter  is  d  is  transformed  into  an 
equivalent  cylinder  having  the  same  diameter ;    find  the 
height  h  of  the  cylinder. 

13.  A  bullet  3  in.  in  diameter  is  recast  in  the  form  of  a 
cylinder  2  in.  in  diameter ;  find  the  height  of  this  cylinder. 

14.  Find  the  height  of  a  right  cone  equivalent  to  a 
hemisphere  8  ft.  in  diameter,  and  having  the  same  base  as 
that  of  the  hemisphere. 

15.  A  sphere  and  an  equivalent   right  cylinder  have 
equal  diameters ;    compare  their  total  surfaces. 

16.  The  frustum  of  a  regular  four-sided  pyramid  is  6  ft. 
high,  and  the  sides  of  its  bases  are  5  ft.  and  8  ft.  respec- 
tively.    What  is  the  height  of  an  equivalent  regular  pyra- 
mid whose  base  is  a  square  with  a  side  12  ft.  long? 

17.  The  frustum  of  a  cone  of  revolution  is  5  ft.  high, 
and  the  diameters  of  its  bases  are  2  ft.  and  3  ft.,  respec- 
tively; find  the  height  of  an  equivalent   right   cylinder 
whose  base  is  equal  in  area  to  the  section  of  the  frustum 
made  by  a  plane  parallel  to  its  bases,  and  equidistant  from 
the  bases. 

18.  The  radii  of  two  spheres  are  r,  r1.     Find  the  radius 
of  a  sphere  equivalent  to  both  the  spheres. 

19.  100  bullets,  equal  in  size,  are  thrown  into  a  cylin- 
drical vessel  8  in.  in  diameter  containing  water.     If  the 
level  of  the  water  rises  4|-  in.,  find  the  diameter  of  a  bullet. 


GEOMETRY.  29 


§  9.   SIMILAR  SOLIDS. 

1.  The  edges  of  two  cubes  are  6  in.  and  7  in. ;  what  is 
the  ratio  of  their  surfaces  ?  their  volumes  ? 

2.  If  the  edge  of  a  cube  is  a,  what  is  the  edge  of  a  cube 
twice  as  large  ? 

3.  The  dimensions  of  a  trunk  are  4  ft.,  3  ft.,  2  ft.     What 
are  the  dimensions  of  a  trunk  similar  in  shape  that  will 
hold  4  times  as  much  ? 

4.  The  diameters  of  two  spheres  are  5  in.  and  20  in., 
respectively.     What  is  the  ratio  of  their  surfaces?  their 
volumes  ? 

5.  The  radii  of  the  earth,  the  moon,  and  the  sun  are  as 
1 :  T\:  112.     If  the  volume  of  the  earth  is  taken  as  unity, 
what  will  be  the  volumes  of  the  moon  and  the  sun  ? 

6.  The  radius  of  a  sphere  is  1  ft.     What  is  the  radius  of 
a  sphere  5  times  as  large  ? 

7.  The  radius  of  a  sphere  is  1  ft.     What  is  the  radius  of 
a  sphere  whose  surface  is  double  that  of  the  given  sphere  ? 

8.  A  sphere  9  in.  in  diameter  is  reduced  in  size  by  a 
lathe  until  the  diameter  is  8  in.     What  part  of  the  whole 
volume  is  removed  ? 

9.  The  diameter  of  an  iron  ball  is  6  in.     What  is  the 
diameter  of  an  iron  ball  weighing  -^  as  much  ? 

10.  Compare  the  surfaces  and  the  volumes  of  two  spheres, 
if  the  diameter  of  one  is  I  that  of  the  other. 

11.  If  a  leaden  ball  1  in.  in  diameter  weighs  T\  of  a 
pound,  what  is  the  diameter  of  a  leaden  ball  that  weighs 
588  pounds? 


30  EXERCISE    MANUAL. 


12,  The  diameter  of  a  globe  is  7  in.     Find  the  diameter 
of  a  globe  whose  volume  is  3  times  that  of  the  first. 

13,  The  diameter  of  a  sphere  is  1  ft.  9  in.     Find  the 
diameter  of  another  sphere  whose  volume  is  i  that  of  the 
former. 

14,  How  much  must  the    dimensions  of  a  cylinder  be 
increased  in  order  to  obtain  a  similar  cylinder   (i.)  whose 
surface  shall  be  n  times  that  of  the  first ;  (ii.)  whose  volume 
shall  be  n  times  that  of  the  first  ? 

15,  A  pyramid  is  cut  by  a  plane  which  passes  midway 
between  the  vertex  and  the  plane  of  the  base.     Compare 
the  volumes  of  the  entire  pyramid  and  the  pyramid  cut  off. 

16,  The  height  of  a  regular  hexagonal  pyramid  is  36  ft., 
and  one  side  of  the  base  is  6  ft.     What  are  the  dimensions 
of  a  similar  pyramid  whose  volume  is  -^  that  of  the  first  ? 

17,  The  length  of  one  of  the  lateral  edges  of  a  pyramid 
is  4m.     How  far  from  the  vertex  will  this  edge  be  cut  by  a 
plane  parallel  to  the  base,  which  divides  the  pyramid  into 
two  equivalent  parts  ? 

18,  The  length  of  a  lateral  edge  of  a  pyramid  is  a.     At 
what  distances  from  the  vertex  will  this  edge  be  cut  by  two 
planes  parallel  to  the  base,  which  divide  the  pyramid  into 
three  equivalent  parts? 

19,  The  length  of  a  lateral  edge  of  a  pyramid  is  a.     At 
what  distance  from  the  vertex  will  this  edge  be  cut  by  a 
plane  parallel  to  the  base,  and  dividing  the  pyramid  into 
two  parts,  which 'are  to  each  other  as  3  :  4? 

20,  The  length  of  a  lateral  edge  of  a  pyramid  is  a.     At 
what  distance  from  the  vertex  must  this  edge  be  cut  by  a 
plane  parallel  to  the  base,  in  order  that  the  volume  of  the 
pyramid  cut  off  may  be  i  of  the  volume  of  the  frustum  ? 


GEOMETRY.  31 


21.  The  volumes  of  two  similar  cones  are  54  cub.  ft.  and 
432  cub.  ft.     The  height  of  the  first  is  6  ft. ;  what  is  the 
height  of  the  other  ? 

22.  The  height  of  a  cone  is  h.     At  what  distance  from 
the  base  must  a  plane  parallel  to  the  base  be  passed,  in 
order  that  the  volume  of  the  cone  cut  off  by  the  plane 
may  be  ij-  of  the  volume  of  the  frustum  which  remains  ? 

23.  The  slant  height  of  a  cone  of  revolution  is  a.     It  is 
required  to  divide  the  cone  by  planes  parallel  to  the  base 
into  three  parts,  the  volumes  of  which  shall  be  proportional 
to  the  numbers  2,  3,  and  5.     At  what  distances  from  the 
vertex  must  the  planes  be  passed  ? 

24.  The  height  of  a  cone  of  revolution  is  h.     At  what 
distances  from  the  vertex  must  planes  parallel  to  the  base 
be  passed,  in  order  that  they  may  divide  the  convex  sur- 
face of  the  cone  into  three  equivalent  parts  ? 

25.  The  height  of  a  cone  of  revolution  is  h,  and  the 
radius  of  its  base  is  r.     What  are  the  dimensions  of  a  simi- 
lar cone  three  times  as  large? 

26.  The  height  of  the  frustum  of  a  right  cone  is  •§-  the 
height  of  the   entire   cone.     Compare  the  volumes  of  the 
frustum  and  the  entire  cone. 

27.  The  frustum  of  a  pyramid  is  8  ft.  high,  and  two 
homologous  edges  of  its  bases  are  4  ft.  and  3  ft.,  respec- 
tively.    Compare  the  volume  of  the  frustum  and  that  of 
the  entire  pyramid. 

28.  In  each  of  two  right  cylinders  the  diameter  is  equal 
to  the  height.     The  volume  of  one  is  f  that  of  the  other. 
What  is  the  ratio  of  their  heights  ? 

29.  Find  the  dimensions  of  a  right  cylinder  ^|  as  large 
as  a  similar  cylinder  whose  height  is  20  ft.,  and  diameter 
10  ft. 


32  EXERCISE   MANUAL. 


30.  If  a  box  will  hold  1200  balls  1  in.  in  diameter,  how 
many  balls  2  in.  in  diameter  will  it  hold  ? 

31.  The  interior   diameter  of   a   hollow   hemispherical 
dome  is  20  ft.,  and  the  thickness  of  the  dome  at  all  points 
is  1  ft.     If  it  costs  $5.00  to  paint  the  interior  surface,  what 
will  it  cost  at  the  same  rate  per  square  foot  to  paint  the 
exterior  surface  ? 

32.  The  exterior  diameter  of  a  spherical  shell  is  20cm, 
and  its  weight  is  ^  that  of  a  solid  ball  made  of  the  same 
material  and  having  the  same  diameter.     Find  the  thick- 
ness of  the  shell. 


§  10.     SOLIDS  OP  REVOLUTION. 

1.  The  sides  of  a  rectangle  are  as  2:1.     What  is  the 
ratio  of  the  volumes  of  the  solids  generated  by  the  revo- 
lution of  the  rectangle  about  these  sides,  respectively,  as 
axes? 

2.  The  sides  of  a  right  triangle  are  3,  4,  5.     Compare 
the  volumes  of  the  solids  generated  by  revolving  the  tri- 
angle about  these  three  sides  respectively. 

3.  Compare  the  volumes  of  the  solids  generated  by  re- 
volving an  isosceles  right  triangle  about  one  of  its  legs  and 
about  its  hypotenuse. 

4.  Compare  the  areas  of  the  curved  surfaces  generated 
by  revolving  an  isosceles  right  triangle  about  one  of  its  legs 
and  about  its  hypotenuse. 

5.  Having  given  the  legs  a,  b  of  a  right  triangle,  find 
the  volume  of  the  solid  generated  by  revolving  the  tri- 
angle about  its  hypotenuse. 

6.  Having  given  the  legs  a,  b  of  a  right  triangle,  find 
the  area  of  the  surface  generated  by  these  two  legs  when 
the  triangle  revolves  about  its  hypotenuse. 


GEOMETRY.  33 


7.  An  equilateral  triangle,  whose  side  is  a,  turns  about 
one  of  its  sides;  find  the  surface  and  the  volume  of  the 
solid  which  is  generated. 

8.  An  equilateral   triangle,  whose   side   is   a,  revolves 
about  an  axis  parallel  to  one  mde  and  passing  through  the 
opposite  vertex.     Find  the  volume  of  the  solid  generated. 

9.  The  base  of  an  isosceles  triangle  =  6  ft.,  and  one  of 
the  equal  sides  =  9  ft.     If  the  triangle  turn  about  an  axis 
passing  through  the  vertex  and  parallel  to  the  base,  find 
the  volume  of  the  solid  which  is  generated. 

10.  Find  the  area  of  the  curved  surface  generated  by  a 
line  A13,  5m  long,  turning  about  an  axis  in  the  same  plane, 
the  -  distances  of  A,  B   from   the  axis  being  3m  and  4m, 
respectively. 

11.  Find  the  volume  of  the  solid  generated  by  an  equi- 
lateral triangle,  whose  side  =  a,  turning  about  an  axis  per- 
pendicular to  one  of  its  sides,  and  passing  through  one  end 
of  this  side. 

12.  One  side  of  an  equilateral  triangle  is  produced  by 
its  own  length,  and   at  the   extremity  a  perpendicular  is 
erected.     If  the  length  of  the  side  is  a,  find  the  volume  of 
the  solid  formed  by  revolving  the  triangle  about  this  per- 
pendicular as  an  axis. 

13.  The  side  of  an  equilateral  triangle  =  a.     The  tri- 
angle is  revolved  about  an  axis  parallel  to  one  side  and  at 
the  distance  b  from  this  side.     Find  the  surface  and  the 
volume  of  the  solid  which  is  generated. 

14.  Having  given  the  hypotenuse  a  of  a  right  triangle, 
find  the  two  legs,  if  the  volumes  of  the  two  cones,  gener- 
ated by  revolving  the  triangle  about  the  legs,  are  as  2:  1. 

15.  Find  the  volume  of  the  solid  generated  by  a  trian- 
gle whose   sides  are  2m,  3m,  and  4m,  and  which  revolves 
about  its  longest  side. 


34  EXERCISE   MANUAL. 


16.  The  radius  of  a  circle  =  2  ft.     Through  a  point  A, 
3  ft.  from  the  centre  of  the  circle,  two  tangents  are  drawn, 
touching  the  circle  in  J9,  (7,  respectively.     Find  the  vol- 
ume of  the  solid  generated  by  revolving  the  triangle  ABC 
about  the  diameter  parallel  to  BO. 

17.  At  the  middle  point  M  of  the  radius  OA  of  a  circle, 
a  perpendicular  is  erected  meeting  the  circumference  in  JV. 
Through  _ZV  a  tangent  is  drawn  meeting  OA  produced  in  P. 
Find   the  volume   of  the  solid  generated  by  the  triangle 
MNP  turning  about  OP  as  an   axis,  the   radius  of  the 
circle  being  r. 

18.  Find   the   surface  and  the  volume  generated  by  a 
square  whose   side  is  a,  revolving  about  an   axis   passing 
through  one  end  of  a  diagonal  of  the  square,  and  perpen- 
dicular to  this  diagonal. 

19.  A  rectangle  ABCD  revolves  about  an  axis  parallel 
to  AB  and  situated  without  the  rectangle.    Prove  that  the 
volume  of  the  solid  generated  is  equal  to  the  area  of  the 
rectangle  multiplied   by  the    circumference   described   by 
the  intersection  of  its  two  diagonals. 

20.  Compare  the  volumes  of  the  solids  generated  by  a 
parallelogram  turning    successively   about    two    adjacent 
sides,  the  lengths  of  these  sides  being  a  and  b. 

21.  A  regular  hexagon,  whose  side  is  a,  turns  about  one 
of  its   sides.     Find   the   surface   and   the  volume  of  the 
solid  which  is  generated. 

22.  A  circle  is  circumscribed  about  a  regular  hexagon 
whose  side  is  a,  and  then  revolved  about  a  diameter  pass- 
ing through  two  opposite  vertices  of  the  hexagon.     Com- 
pare the  volumes  of  the  solids  generated  by  the  circle  and 
the  hexagon. 


GEOMETRY.  35 


23.  A  circle  is  circumscribed  about  a  regular  hexagon 
whose  side  is  a,  and  then  revolved  about  a  diameter  pass- 
ing through  the  middle  points  of  two  opposite  sides  of  the 
hexagon.      Compare  the  volumes  of  the  solids  generated 
by  the  circle  and  the  hexagon. 

24.  What  must  be  the  angle  of  a  circular  sector  AOC, 
in  order  that  the  volume  of  the  spherical  sector,  generated 
by  the  revolution  of  the  circular  sector  about  AO,  may  be 
exactly  i  of  the  entire  volume  of  the  sphere  to  which  it 
belongs  ? 

25.  The  diameter  of  a  circle  is  4  ft.     Find  the  surface 
and  the  volume  of  the  solid  generated  by  a  chord  2  ft. 
long,  revolving  about  a  diameter  parallel  to  the  chord. 

26.  Upon  the  diameter  AOB  of  a  semicircle,  of  which 
0  is  the  centre  and  r  the  radius,  semicircles  are  described 
having   for  their  diameters  AO  and  OB.     The  figure  is 
then  revolved  about  AOB  as  an  axis.     Find  the  volume  of 
the  solid  generated  by  the  surface  comprised  between  the 
three  semicircles. 

27.  An  equilateral  triangle  revolves  about  one   of  its 
altitudes.     Compare  the  convex  surface  of  the  cone  gen- 
erated by  the  triangle  and  the  surface  of  the  sphere  gen- 
erated by  the  circle  inscribed  in  the  triangle. 

28.  An   equilateral  triangle  revolves  about  one   of  its 
altitudes.    Compare  the  volumes  of  the  solids  generated  by 
the  triangle,  the  inscribed   circle,  and  the  circumscribed 
circle. 

29.  A  circle  is  inscribed  in  a  square,  and  the  figure  is 
revolved  about  the  diameter,  passing  through  two  opposite 
points  of  contact.      Compare  the  convex   surface   of  the 
cylinder  generated  by  the  square,  and  the  surface  of  the 
sphere  generated  by  the  circle. 


36  EXERCISE    MANUAL. 

30.  A  circle  is  inscribed  in  a  square,  and  one  of  the 
points  of  contact  is  joined  to  the  extremities  of  the  oppo- 
site side  of  the  square.  The  figure  is  then  revolved  about 
the  diameter  which  passes  through  this  point  of  contact. 
Compare  the  volumes  of  the  cylinder,  sphere,  and  cone 
which  are  generated. 

§  11.    INSCRIBED  AND  CIRCUMSCRIBED  SOLIDS. 

1.  Find  the  volume  in  cubic  feet  of  the  largest  square 
beam  which  can  be  made  from  a  log  in  the  shape  of  a  right 
cylinder  40  ft.  long  and  2  ft.  in  diameter. 

2.  The  edge  of  a  cube  is  14  ft.     What  is  the  convex 
surface  of  the  largest  cylinder  that  can  be  made  from  it? 

3.  Compare  the  volumes  of  the  largest  cylinder,  sphere, 
and  cone  which  can  be  made  from  a  given  cube. 

4.  Compare  the  volumes  of  the  largest  cylinder,  sphere, 
and  cone  which  can  be  made  from  a  right  prism  a  ft.  long, 
the  base  of  which  is  a  square  with  a  side  b  ft.  long. 

5.  What  is  the  ratio  of  the  volume  of  a  regular  quad- 
rangular pyramid  to  that  of  the  inscribed  cone  ? 

6.  What  is  the  ratio  of  the  volume  of  a  cone  of  revolu- 
tion to  that  of  the  inscribed  regular  quadrangular  pyramid  ? 

7.  If  a  wooden  cube  weighs  21  Ibs.,  what  will  be  the 
weight  of  the  largest  cone  which  can  be  turned  from  it  ? 

8.  The  edge  of  a  regular  tetrahedron  is  a.     Find  the 
radii  r,  r'  of  the  inscribed  and  circumscribed  spheres,  the 
surface  JS,  and  the  volume  V. 

9.  The  same  exercise  for  the  regular  octahedron. 

10.  Compare   the  volume   of  the   cube   inscribed  in  a 
given  sphere  with  that  of  the  cube  circumscribed  about  the 
same  sphere. 


GEOMETRY.  37 


11.  The  same  exercise  for  the  regular  tetrahedron. 

12.  The  same  exercise  for  the  regular  octahedron. 

13.  The  height  of  a  frustum  of  a  right  cone  is  A,  and 
the   radii   of  its  bases  r,  r';  what  is  the  volume  of  the 
largest   regular  four-sided    frustum    which  can   be   made 
from  it  ? 

14.  The  volume  of  a  right  cone,  whose  slant  height  is 
equal  to  the  diameter  of  its  base,  is  F;  find  the  volume  of 
the  circumscribed  regular  triangular  pyramid. 

15.  A  cone,  whose  slant  height  is  equal  to  the  diameter 
of  its  base,  is   inscribed  in  a  given  sphere,  and  a  similar 
cone   is  circumscribed  about  the  same  sphere.     Compare 
the  volumes  of  these  two  cones. 

16.  The  radius  of  the  base  of  a  right  cone  is  a,  and  the 
radius  of  the  inscribed  sphere  is  b ;  find  the  volume  of  the 
cone. 

17.  The  volume  of  the  frustum  of  a  right  cone  is  F;  find 
the  difference  between  the  volumes  of  the  circumscribed 
and  the  inscribed  regular  hexagonal  frustums. 

18.  The  volume  of  the  frustum  of  a  regular  quandran- 
gular  pyramid  is  F;   find  the  difference  between  the  vol- 
umes of  the  circumscribed  and  the  inscribed  frustums  of 
cones. 

19.  The  height  of  a  cone  circumscribed  about  a  sphere 
i»  double  the  diameter  of  the  sphere.     If  the  surface  of  the 
sphere  is  8,  find  the  total  surface  of  the  cone. 

20.  Find  the  height  of  a  right  cone  inscribed  in  a  sphere 
of  radius  r,  if  the  convex  surface  of  the  cone  is  equal  to 
twice  the  area  of  its  base. 


38  EXERCISE    MANUAL. 

USEFUL  FORMULA. 

V  stands  for  volume.  S  stands  for  the  lateral  surface  of  the  right 
prism,  regular  pyramid,  or  frustum  of  regular  pyramid,  and  the  con- 
vex surface  of  the  cylinder  of  revolution,  cone  of  revolution,  or  frus- 
tum of  cone  of  revolution  ;  T  stands  for  the  total  surface  of  each  of 
these  solids,  respectively. 

The  Prism. 

S  =  perimeter  of  base  X  height. 
T  =  8  +  twice  area  of  base. 
V  =  area  of  base  X  height. 

The  Cylinder  of  Revolution. 
Radius  of  base,  r  ;  height,  A. 


T=  Z 
V= 


The  Pyramid. 

8  =  i  perimeter  of  base  X  slant  height. 

T  =  S  -j-  area  of  base. 

V  =  i  area  of  base  X  height. 

The  Cone  of  Revolution. 
Radius  of  base,  r  ;  height,  h  ;  slant  height,  a. 

7T7-2. 


Frustum  of  Pyramid. 

Perimeters  of  bases,  p  and  p'  ;  areas  of  bases,  B  and  B'  ;  height,  h. 
S=  $  (p+p'}  X  slant  height. 
T=S+  B  +  B'. 


GEOMETRY.  39 


Frustum  of  Cone  of  Revolution. 
Radii  of  bases,  r  and  ?•'  ;  height,  h  ;  slant  height,  a. 


The  Sphere. 

Radius,  r  ;  angle  of  lune,  o  ;  angles  of  spherical  A,  o,  j8,  y  ; 
height  of  zone  or  segment,  h. 

Surface  of  sphere       =  47rr2. 

Area  of  lune 

90 

Area  of  spherical  A  =  <>  +  /*  +  T  Q~  180°>  X  4^. 
Area  of  zone  =  2?rrA. 


Vol.  of  sphere 

Vol.  of  wedge  =  |r  X  area  of  base. 

Vol.  of  sph.  pyramid  =  £?•  X  area  of  base. 

Vol.  of  sph.  sector     =  1  7rr2A. 

Vol.  of  sph.  segment=  J  h  (sum  of  bases)  -f- 


USEFUL  NUMBERS. 

V2"=  1.41421.  -v/2~=  1.25992. 

V3"=  1.73205.  ^3"=  1.44224. 

V5~=  2.23606.  \/5"=  1.70998. 

V7"=  2.64575.  ^7"=  1.91293. 


ANSWEKS 

TO 

EXERCISES   IN   SOLID    GEOMETRY. 


§  1.     PLANES. 

10. 

12  in.               13. 

12  in. 

17.    5.143  in.,  6.857  in. 

11. 

10  in.               14. 

20.44  in. 

33.    2.828  in. 

12. 

7.071  in.         15. 

24°,  156°. 

34.    60°. 

§ 

2.     THE  PRISM. 

1. 

6  dl  ;  a?  ;  a  V3. 

18. 

2(ab  +  ac  +  be)  ] 

2. 

4  ft.  ;  64  cub.  ft. 

abc  ;   Va2  -f-  £2  +  c2. 

3. 

J#    1     P 
\6'  6\6' 
10  ft.  ;  600  sq.  ft. 

19. 
20. 

3pF      3pF      'W 

\  np       \  wip      \  mn 

4. 

20.        ' 

21 

S==4:ak;V=a?h. 

5. 

Each  edge  =  lm. 

22. 

£_6aA.'-pr__3aW3 

fl 

a^/2 

2 

A 

23. 

<>  Q   '  .  17—  a2h(  l~f~  v  2") 

7. 

405  sq.  ft. 

24. 

4 

168  cub  in. 

8. 

512  cub.  ft. 

25. 

31  cub.  ft.  864  cub.  in. 

9. 

4ft. 

26. 

792  cub.  in.;  1440cub.in. 

10. 

2250  Ibs. 

27. 

18  cub.  ft.  73.2  cub.  in. 

11. 

82.752kg. 

28. 

1920  cub.  ft. 

12. 

62  sq.  ft.  24  sq.  in.             29    94^^ 

13. 

15,360  bricks. 

30. 

187.0632  cub.  in. 

14. 

1  ft.  8  in. 

31. 

140,000  cub.  ft. 

15. 
16. 

4145  Ibs.  13  oz.  ; 
6255  Ibs.  3oz. 
1  cub.  ft.  270  sub 

32. 

33. 

.  in. 

117,333,333*  cub.  yds. 
1558.8  cub.  ft. 

17. 

1815  cub.  ft 

34. 

i  2y 

42 


ANSWERS. 


§  3.     THE  CYLINDER  OF  REVOLUTION. 


7. 
8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 


251.4  sq.  ft. 

$=73  sq.  ft.  48  sq.  in.  ; 

V=  42  cub.  ft.  1344  cub.  in. 

Tin. 


12.53 


n. 


4,  2,  and  8  times,  respec- 

tively. 
$31.43. 

1746^  cub.  yds. 
2887|  gals. 
7392yds. 
294  Ibs.  10-f-  oz. 
154  sq.in.;  2156  cub.  in. 
170.903kg. 


2.26 


18.  1.768kg. 

19.  1  cub.  ft.  1157|  cub.  in. 

20.  $165;  $12.83. 

21.  1  ft.  3tf  in. 

22.  16  ft.  6.88  in. 

23.  1540  revolutions. 

24.  2346|  ft. 

25.  95  Ibs.  12  oz. 

26.  92491. 

27.  8482.32ccm. 

28.  546.89kg. 


31.  V^- 


§  4.     THE  PYRAMID. 

1.  105  sq.  ft.  6.  25.85  ft. 

2.  11.54  ft.  7.  33i  cub.  ft. 

3.  6  ft.  8.  34.64  cub.  ft. 

4.  7.11  sq.  ft.  9.  935.316  cub.  ft. 

5.  14.87  sq.  ft. ;  23.38  sq.  ft.  10.  384  sq.  ft. 


GEOMETRY. 


43 


11.  177.588  sq.  ft. 

12.  5096  sq.  ft. 

13.  Gift. 

14.  10ft. 

15.  51,710.933  cub.  ft. 

16.  266.29  sq.  ft. 

17.  Slant  height,  12cm; 
area  of  one  face,  108qc 
height,  10.816C">; 
total  surface,  464.29qci 
volume,  505.85cbm. 


18.  185  sq.  ft, 

19.  198  oz.  Avoir. 

20.  519  cub.  ft. 


21. 


22.  A=-- 


24.    - 


§  5.    THE  CONE  OF  REVOLUTION. 


1.  0.16qm. 

2.  6ft. 

3.  2.309  ft. 

4.  Area  of  section  = 

5.  1.414  ft. 


6. 

7.  12ft. 

8.  28  ft. 

9.  3  cub.  ft.  llli  cub.  in. 

10.  $351. 

11.  96.052  sq.  ft. 

12.  38^  sq.ft. 

13.  $67.13. 

14.  27ff  Yds- 

15.  fTra2. 

16.  2F;  4F;  8F 

17.  103,488  cub.  yds. 

18.  18.935ft. 


19.  77ft  cub.  ft. 

20.  34  cub.  ft.  310£  cub.  in. 

21.  llGVft- 

22.  2256||  gals. 

23.  3|  cub.  in. 

24.  5&cub.ft. 

25.  r  =  3.85cm; 
h  =  15.4cm. 

26.  21i  cub.  ft. 

27.  3.2969qm. 

28.  4.7952om. 

29.  1982.4ccm. 

7    W 
9 


30.  ±+£- 


81. 

o 


44  ANSWERS. 


§  6.    FRUSTUMS  OF  PYRAMIDS  AND  CONES. 

1. 

2. 
3. 
4. 

11  sq.  ft.  66  sq.  in. 
430.3924  sq.  ft, 
$132. 
40  *•  =  126.  3636*". 

14. 
15. 
16. 

12fJ  gals. 
515  cub.  ft. 
1.98  ft. 
28.421cbm. 

5. 
6. 

7. 

8. 
9. 

5.28m. 
34.22  sq.  ft 

17. 
18. 
19. 
20. 
21. 
22. 

356.2qm;  214.282cbm. 
0.2816m. 
6  ft.  and  3  ft. 
144  sq.ft.;    78.83  cub.  ft. 
1693i£  cub.  in. 
716.2848cbm. 

\of-\-ar 
123.41  cub.  ft. 

10. 
11. 

1064  cub.  ft. 
13,312  cub.  ft. 

23. 

9A 

22,140  cub.  ft. 

12.  $35.45. 

§7.     THE  SPHERE. 

2.  12.06qm.  11.  $1140.48. 

3.  1.74m.  12.  4& 

4.  11  ft. ;  9.898  ft.  13.  0=  VS& 

5.  18.09qm.  14.  £. 

6.  3.69  in.  15.  £  sq.  ft. 

7.  (i.)  2  sq.  ft.  26£  sq.  in. ;  16.  40°. 
(ii.)  9  sq.  ft.  90  sq.  in. ;  17.  TV 

(iii.)  17  sq.  ft.  16  sq.  in.  ;  18.  102|  sq.  ft. 

(iv.)  154  sq.ft.;  19.  410|  sq.  ft. 

(v.)  55.44  sq.  ft. ;  20.  |. 

(vi.)  346.5  sq.  ft.  21.  51£  sq.  ft. 

8.  (i.)  1  ft.  2  in. ;  22.  8  ft. 

(ii.)  3  ft.  6  in. ;  23.  1386  sq.  ft. 

(iii.)  56  ft.  24.  308 sq.ft. 

9f  693  sq.  ft.  25.  2772  sq.  ft. 

10.  136|  sq.ft.  26.  6ft. 


GEOMETRY.  45 


27.  330  sq.  ft. 

50.   66fi  cub.  in. 

28.  7r(A2-f  r2);  27r(A2-j-r2). 

51.   10^-|-|-  gals. 

29    h--. 

52.  18  Ibs.  6T9^  oz. 

~2" 

53.  7  in. 

30.  Area  of  zone,  25|  sq.  ft.  ; 
areas  of  bases,  22  sq.  ft. 

54.   3-f-g-,  nearly. 
55.  5.46  in. 

and  37f  sq.  ft. 

31.  Radius  of  circle  =V2rA. 

57.  54|f  Ibs. 

32.   10  sq.  ft.  131f  sq.  in. 

33.  213.714  sq.  ft. 
34.  1977.1  sq.  ft. 

59.   143ft  cub.  ft. 
60.  60°. 

35.  80,457,143  sq.mi.,  nearly. 

61.  1  cub.  ft. 

36.  r(V5~-2). 

37.  r(V2~-l). 

'         3 

38.  7rr2(2-V3). 

£Q      "yy  *-)\s  TTtb 

13 

39    2?rr2A 

64.  239fcub.ft. 

40.  i. 

65.  179|cub.ft. 

41.  ^  the  radius. 

66.  22ii  cub.  ft. 

42.  5.03167  miles. 

67.  54,476^-  cub.  ft. 

43.  (i.)  1437i  cub.  in.  ; 

rtQ           £\t   g  r,             , 

(ii.)  22H  cub.  ft.  ; 

(iii.)  606|  cub.  ft.  ; 

70.  h=      S      • 

(iv.)  2807^  cub.  ft.  ; 

VWF 

(v.)  1663.893  cub.  ft.  ; 

71.  485f  cub.ft. 

(vi.)  38,808  cub.  ft. 

72.  234|  cub.  ft. 

44.  (i.)  5ft.3in.;(iii.)12.6ft.; 

73.  19fMfgalB. 

(ii.)  7ft.;       (iv.)  4.2  yds. 

74.  205Jcub.ft. 

45.  1538.8  cub.  ft. 

75.  654.498cbm. 

46.  V=~ 

76.   1:3;   1  :  27. 

77       7          i       /o            /R\ 

47.  r  =  Ji-£ 

78.  F=^(r2-r'2). 

79    r=¥- 

48.  r  =  W6  =  1.2247. 

TTfl           O 

49.  0.665cbm  80. 


46  ANSWERS. 


§  8.     EQUIVALENT  SOLIDS. 

1.  5:4. 

.8. 

k-~ 

2.  9cm;   24q?m. 

Trb* 

Z.                        Q  7 

9. 

3  ft. 

3.   (i.)_  .  ;    (ii.)  —  —  ; 

10. 

4ft. 

TTOt,                        TTd 

11. 

3.629  ft. 

/•••   \        \OdOC 

19 

7>      -  2  fl 

\     4lT 

J-£J  • 

13. 

n      -g-a. 
4^  in. 

4     3|3< 

14. 

8ft. 

\  4:7T 

15. 

6:7. 

5      3|4?rr3 

16. 

5f  ft. 

\    3 

17. 

5.0666  ft. 

6.  4  ft. 

18. 

•v^?+V*. 

7.  77ft. 

19. 

1.5869  in. 

§9. 

SIMILAR 

SOLIDS. 

1.  36:49;    216:343. 

15. 

8:1. 

2.  a  A/2. 

16. 

Height,  13.2625  ft.  ; 

3.  6.35ft.;  4.76ft.;  3 

.17ft. 

side  of  base,  2.2104  ft. 

4.   1:16;    1:64. 

17. 

3.17m. 

5.  yf-L;    1,404,928. 

18. 

«A/X  and  aA/F 

6.  A/5. 

19. 

1/5- 
avf. 

7.  1.414ft. 

20. 

«A/f 

8217 
*     7  2  9  * 

21. 

12ft. 

9.  2  in. 

DO 

A 

&A. 

10.   Surfaces  are  as  25  : 

64; 

2 

volumes  are  as  125 

:512.    23. 

«A/-§-,  and  aA/f 

11.  14  in. 

24 

AV3        T  ^A/6 

12.  10.094  in. 

3                 3    ' 

13.  10*  in. 

25. 

Height  =  ^A/3  ; 

14.  (i.)  A/n  times  ; 

radius  of  base  =  r  A/3". 

(ii.)  Vw  times. 

26. 

98:125.         27.  37:64. 

GEOMETRY.  47 


28.  1  :  0.90856.  30.  150  balls. 

29.  Height,  19.5743;  31.  $6.05. 
diameter,  9.7871.                32.  0.3451cm. 


§  10.    SOLIDS  OF  REVOLUTION. 

1-   1:2.  aV5        , 

2.20:15:12.  ~5~ 


3.  2:V2. 

4.  1:1.  16.  3407rV^cub.ft. 

81 

17.  F= 


=ira»  V2. 


22.4:3. 


25.  fl=  47rV3  sq.  ft.  ; 
F=6ircub.  ft. 

26.  F=7rr3. 

27.  3  :  2. 

4  28.  9:4:32. 

13.  S  =  7ra  (a  V3  -h  6  i)  ;          29.  They  are  equal. 

30.  3:2:1. 


§  11.     INSCRIBED  AND  CIRCUMSCRIBED  SOLIDS. 

1.  80  cub.  ft.  4.  3a:  2b  :  a. 

2.  616  sq.  ft.  5.   14  :  11. 

3.  3  :  2  :  1.  6.  11 :  7. 


48  ANSWERS. 


7.  5J  Ibs.  13. 

~~12  ~4 

~12 

a  a  16- 

'  7  —Q       >  7  — 2 

17. 


10.  V3:9.  18.  i^F. 

11.  1:8.  19.  2& 

12.  1 :  2V2.  20.  rV3. 


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MAP     OA    IQjfjf 

svJHr     o\j    IJJ44 

SEP  1U  1947 

'w; 

TT?  3  10  Wi  it 

n        ffffil    A/ 

4Sep56lWi 

REC'D  LD 

crn       A    inrr* 

oLr     4  lyob 

4  May  590  fH 

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